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java.io.FileNotFoundException: data.csv (The system cannot find the file specified) [duplicate]

I want to access my current working directory using Java.
My code:
String currentPath = new java.io.File(".").getCanonicalPath();
System.out.println("Current dir:" + currentPath);
String currentDir = System.getProperty("user.dir");
System.out.println("Current dir using System:" + currentDir);
Output:
Current dir: C:\WINDOWS\system32
Current dir using System: C:\WINDOWS\system32
My output is not correct because the C drive is not my current directory.
How to get the current directory?

Code :
public class JavaApplication {
public static void main(String[] args) {
System.out.println("Working Directory = " + System.getProperty("user.dir"));
}
}
This will print the absolute path of the current directory from where your application was initialized.
Explanation:
From the documentation:
java.io package resolve relative pathnames using current user directory. The current directory is represented as system property, that is, user.dir and is the directory from where the JVM was invoked.

See: Path Operations (The Java™ Tutorials > Essential Classes > Basic I/O).
Using java.nio.file.Path and java.nio.file.Paths, you can do the following to show what Java thinks is your current path. This for 7 and on, and uses NIO.
Path currentRelativePath = Paths.get("");
String s = currentRelativePath.toAbsolutePath().toString();
System.out.println("Current absolute path is: " + s);
This outputs:
Current absolute path is: /Users/george/NetBeansProjects/Tutorials
that in my case is where I ran the class from.
Constructing paths in a relative way, by not using a leading separator to indicate you are constructing an absolute path, will use this relative path as the starting point.

The following works on Java 7 and up (see here for documentation).
import java.nio.file.Paths;
Paths.get(".").toAbsolutePath().normalize().toString();

This will give you the path of your current working directory:
Path path = FileSystems.getDefault().getPath(".");
And this will give you the path to a file called "Foo.txt" in the working directory:
Path path = FileSystems.getDefault().getPath("Foo.txt");
Edit :
To obtain an absolute path of current directory:
Path path = FileSystems.getDefault().getPath(".").toAbsolutePath();
* Update *
To get current working directory:
Path path = FileSystems.getDefault().getPath("").toAbsolutePath();

Java 11 and newer
This solution is better than others and more portable:
Path cwd = Path.of("").toAbsolutePath();
Or even
String cwd = Path.of("").toAbsolutePath().toString();

This is the solution for me
File currentDir = new File("");

What makes you think that c:\windows\system32 is not your current directory? The user.dir property is explicitly to be "User's current working directory".
To put it another way, unless you start Java from the command line, c:\windows\system32 probably is your CWD. That is, if you are double-clicking to start your program, the CWD is unlikely to be the directory that you are double clicking from.
Edit: It appears that this is only true for old windows and/or Java versions.

Use CodeSource#getLocation().
This works fine in JAR files as well. You can obtain CodeSource by ProtectionDomain#getCodeSource() and the ProtectionDomain in turn can be obtained by Class#getProtectionDomain().
public class Test {
public static void main(String... args) throws Exception {
URL location = Test.class.getProtectionDomain().getCodeSource().getLocation();
System.out.println(location.getFile());
}
}

this.getClass().getClassLoader().getResource("").getPath()

generally, as a File object:
File getCwd() {
return new File("").getAbsoluteFile();
}
you may want to have full qualified string like "D:/a/b/c" doing:
getCwd().getAbsolutePath()

I'm on Linux and get same result for both of these approaches:
#Test
public void aaa()
{
System.err.println(Paths.get("").toAbsolutePath().toString());
System.err.println(System.getProperty("user.dir"));
}
Paths.get("") docs
System.getProperty("user.dir") docs

I hope you want to access the current directory including the package i.e. If your Java program is in c:\myApp\com\foo\src\service\MyTest.java and you want to print until c:\myApp\com\foo\src\service then you can try the following code:
String myCurrentDir = System.getProperty("user.dir")
+ File.separator
+ System.getProperty("sun.java.command")
.substring(0, System.getProperty("sun.java.command").lastIndexOf("."))
.replace(".", File.separator);
System.out.println(myCurrentDir);
Note: This code is only tested in Windows with Oracle JRE.

On Linux when you run a jar file from terminal, these both will return the same String: "/home/CurrentUser", no matter, where youre jar file is. It depends just on what current directory are you using with your terminal, when you start the jar file.
Paths.get("").toAbsolutePath().toString();
System.getProperty("user.dir");
If your Class with main would be called MainClass, then try:
MainClass.class.getProtectionDomain().getCodeSource().getLocation().getFile();
This will return a String with absolute path of the jar file.

Using Windows user.dir returns the directory as expected, but NOT when you start your application with elevated rights (run as admin), in that case you get C:\WINDOWS\system32

Mention that it is checked only in Windows but i think it works perfect on other Operating Systems [Linux,MacOs,Solaris] :).
I had 2 .jar files in the same directory . I wanted from the one .jar file to start the other .jar file which is in the same directory.
The problem is that when you start it from the cmd the current directory is system32.
Warnings!
The below seems to work pretty well in all the test i have done even
with folder name ;][[;'57f2g34g87-8+9-09!2##!$%^^&() or ()%&$%^##
it works well.
I am using the ProcessBuilder with the below as following:
🍂..
//The class from which i called this was the class `Main`
String path = getBasePathForClass(Main.class);
String applicationPath= new File(path + "application.jar").getAbsolutePath();
System.out.println("Directory Path is : "+applicationPath);
//Your know try catch here
//Mention that sometimes it doesn't work for example with folder `;][[;'57f2g34g87-8+9-09!2##!$%^^&()`
ProcessBuilder builder = new ProcessBuilder("java", "-jar", applicationPath);
builder.redirectErrorStream(true);
Process process = builder.start();
//...code
🍂getBasePathForClass(Class<?> classs):
/**
* Returns the absolute path of the current directory in which the given
* class
* file is.
*
* #param classs
* #return The absolute path of the current directory in which the class
* file is.
* #author GOXR3PLUS[StackOverFlow user] + bachden [StackOverFlow user]
*/
public static final String getBasePathForClass(Class<?> classs) {
// Local variables
File file;
String basePath = "";
boolean failed = false;
// Let's give a first try
try {
file = new File(classs.getProtectionDomain().getCodeSource().getLocation().toURI().getPath());
if (file.isFile() || file.getPath().endsWith(".jar") || file.getPath().endsWith(".zip")) {
basePath = file.getParent();
} else {
basePath = file.getPath();
}
} catch (URISyntaxException ex) {
failed = true;
Logger.getLogger(classs.getName()).log(Level.WARNING,
"Cannot firgue out base path for class with way (1): ", ex);
}
// The above failed?
if (failed) {
try {
file = new File(classs.getClassLoader().getResource("").toURI().getPath());
basePath = file.getAbsolutePath();
// the below is for testing purposes...
// starts with File.separator?
// String l = local.replaceFirst("[" + File.separator +
// "/\\\\]", "")
} catch (URISyntaxException ex) {
Logger.getLogger(classs.getName()).log(Level.WARNING,
"Cannot firgue out base path for class with way (2): ", ex);
}
}
// fix to run inside eclipse
if (basePath.endsWith(File.separator + "lib") || basePath.endsWith(File.separator + "bin")
|| basePath.endsWith("bin" + File.separator) || basePath.endsWith("lib" + File.separator)) {
basePath = basePath.substring(0, basePath.length() - 4);
}
// fix to run inside netbeans
if (basePath.endsWith(File.separator + "build" + File.separator + "classes")) {
basePath = basePath.substring(0, basePath.length() - 14);
}
// end fix
if (!basePath.endsWith(File.separator)) {
basePath = basePath + File.separator;
}
return basePath;
}

assume that you're trying to run your project inside eclipse, or netbean or stand alone from command line. I have write a method to fix it
public static final String getBasePathForClass(Class<?> clazz) {
File file;
try {
String basePath = null;
file = new File(clazz.getProtectionDomain().getCodeSource().getLocation().toURI().getPath());
if (file.isFile() || file.getPath().endsWith(".jar") || file.getPath().endsWith(".zip")) {
basePath = file.getParent();
} else {
basePath = file.getPath();
}
// fix to run inside eclipse
if (basePath.endsWith(File.separator + "lib") || basePath.endsWith(File.separator + "bin")
|| basePath.endsWith("bin" + File.separator) || basePath.endsWith("lib" + File.separator)) {
basePath = basePath.substring(0, basePath.length() - 4);
}
// fix to run inside netbean
if (basePath.endsWith(File.separator + "build" + File.separator + "classes")) {
basePath = basePath.substring(0, basePath.length() - 14);
}
// end fix
if (!basePath.endsWith(File.separator)) {
basePath = basePath + File.separator;
}
return basePath;
} catch (URISyntaxException e) {
throw new RuntimeException("Cannot firgue out base path for class: " + clazz.getName());
}
}
To use, everywhere you want to get base path to read file, you can pass your anchor class to above method, result may be the thing you need :D
Best,

For Java 11 you could also use:
var path = Path.of(".").toRealPath();

This is a very confuse topic, and we need to understand some concepts before providing a real solution.
The File, and NIO File Api approaches with relative paths "" or "." uses internally the system parameter "user.dir" value to determine the return location.
The "user.dir" value is based on the USER working directory, and the behavior of that value depends on the operative system, and the way the jar is executed.
For example, executing a JAR from Linux using a File Explorer (opening it by double click) will set user.dir with the user home directory, regardless of the location of the jar. If the same jar is executed from command line, it will return the jar location, because each cd command to the jar location modified the working directory.
Having said that, the solutions using Java NIO, Files or "user.dir" property will work for all the scenarios in the way the "user.dir" has the correct value.
String userDirectory = System.getProperty("user.dir");
String userDirectory2 = new File("").getAbsolutePath();
String userDirectory3 = Paths.get("").toAbsolutePath().toString();
We could use the following code:
new File(MyApp.class.getProtectionDomain()
.getCodeSource()
.getLocation()
.toURI().getPath())
.getParent();
to get the current location of the executed JAR, and personally I used the following approach to get the expected location and overriding the "user.dir" system property at the very beginning of the application. So, later when the other approaches are used, I will get the expected values always.
More details here -> https://blog.adamgamboa.dev/getting-current-directory-path-in-java/
public class MyApp {
static {
//This static block runs at the very begin of the APP, even before the main method.
try{
File file = new File(MyApp.class.getProtectionDomain().getCodeSource()
.getLocation().toURI().getPath());
String basePath = file.getParent();
//Overrides the existing value of "user.dir"
System.getProperties().put("user.dir", basePath);
}catch(URISyntaxException ex){
//log the error
}
}
public static void main(String args []){
//Your app logic
//All these approaches should return the expected value
//regardless of the way the jar is executed.
String userDirectory = System.getProperty("user.dir");
String userDirectory2 = new File("").getAbsolutePath();
String userDirectory3 = Paths.get("").toAbsolutePath().toString();
}
}
I hope this explanation and details are helpful to others...

Current working directory is defined differently in different Java implementations. For certain version prior to Java 7 there was no consistent way to get the working directory. You could work around this by launching Java file with -D and defining a variable to hold the info
Something like
java -D com.mycompany.workingDir="%0"
That's not quite right, but you get the idea. Then System.getProperty("com.mycompany.workingDir")...

This is my silver bullet when ever the moment of confusion bubbles in.(Call it as first thing in main). Maybe for example JVM is slipped to be different version by IDE. This static function searches current process PID and opens VisualVM on that pid. Confusion stops right there because you want it all and you get it...
public static void callJVisualVM() {
System.out.println("USER:DIR!:" + System.getProperty("user.dir"));
//next search current jdk/jre
String jre_root = null;
String start = "vir";
try {
java.lang.management.RuntimeMXBean runtime =
java.lang.management.ManagementFactory.getRuntimeMXBean();
String jvmName = runtime.getName();
System.out.println("JVM Name = " + jvmName);
long pid = Long.valueOf(jvmName.split("#")[0]);
System.out.println("JVM PID = " + pid);
Runtime thisRun = Runtime.getRuntime();
jre_root = System.getProperty("java.home");
System.out.println("jre_root:" + jre_root);
start = jre_root.concat("\\..\\bin\\jvisualvm.exe " + "--openpid " + pid);
thisRun.exec(start);
} catch (Exception e) {
System.getProperties().list(System.out);
e.printStackTrace();
}
}

This isn't exactly what's asked, but here's an important note: When running Java on a Windows machine, the Oracle installer puts a "java.exe" into C:\Windows\system32, and this is what acts as the launcher for the Java application (UNLESS there's a java.exe earlier in the PATH, and the Java app is run from the command-line). This is why File(".") keeps returning C:\Windows\system32, and why running examples from macOS or *nix implementations keep coming back with different results from Windows.
Unfortunately, there's really no universally correct answer to this one, as far as I have found in twenty years of Java coding unless you want to create your own native launcher executable using JNI Invocation, and get the current working directory from the native launcher code when it's launched. Everything else is going to have at least some nuance that could break under certain situations.

Try something like this I know I am late for the answer but this obvious thing happened in java8 a new version from where this question is asked but..
The code
import java.io.File;
public class Find_this_dir {
public static void main(String[] args) {
//some sort of a bug in java path is correct but file dose not exist
File this_dir = new File("");
//but these both commands work too to get current dir
// File this_dir_2 = new File(this_dir.getAbsolutePath());
File this_dir_2 = new File(new File("").getAbsolutePath());
System.out.println("new File(" + "\"\"" + ")");
System.out.println(this_dir.getAbsolutePath());
System.out.println(this_dir.exists());
System.out.println("");
System.out.println("new File(" + "new File(" + "\"\"" + ").getAbsolutePath()" + ")");
System.out.println(this_dir_2.getAbsolutePath());
System.out.println(this_dir_2.exists());
}
}
This will work and show you the current path but I don't now why java fails to find current dir in new File(""); besides I am using Java8 compiler...
This works just fine I even tested it new File(new File("").getAbsolutePath());
Now you have current directory in a File object so (Example file object is f then),
f.getAbsolutePath() will give you the path in a String varaible type...
Tested in another directory that is not drive C works fine

My favorite method is to get it from the system environment variables attached to the current running process. In this case, your application is being managed by the JVM.
String currentDir = System.getenv("PWD");
/*
/home/$User/Documents/java
*/
To view other environment variables that you might find useful like, home dir, os version ........
//Home directory
String HomeDir = System.getEnv("HOME");
//Outputs for unix
/home/$USER
//Device user
String user = System.getEnv("USERNAME");
//Outputs for unix
$USER
The beautiful thing with this approach is that all paths will be resolved for all types of OS platform

You might use new File("./"). This way isDirectory() returns true (at least on Windows platform). On the other hand new File("") isDirectory() returns false.

None of the answers posted here worked for me. Here is what did work:
java.nio.file.Paths.get(
getClass().getProtectionDomain().getCodeSource().getLocation().toURI()
);
Edit: The final version in my code:
URL myURL = getClass().getProtectionDomain().getCodeSource().getLocation();
java.net.URI myURI = null;
try {
myURI = myURL.toURI();
} catch (URISyntaxException e1)
{}
return java.nio.file.Paths.get(myURI).toFile().toString()

System.getProperty("java.class.path")

Java: Is there a way, how I can import a file which is in the same folder as the .jar? [duplicate]

Want to improve this post? Provide detailed answers to this question, including citations and an explanation of why your answer is correct. Answers without enough detail may be edited or deleted.
My code runs inside a JAR file, say foo.jar, and I need to know, in the code, in which folder the running foo.jar is.
So, if foo.jar is in C:\FOO\, I want to get that path no matter what my current working directory is.

return new File(MyClass.class.getProtectionDomain().getCodeSource().getLocation()
.toURI()).getPath();
Replace "MyClass" with the name of your class.
Obviously, this will do odd things if your class was loaded from a non-file location.

Best solution for me:
String path = Test.class.getProtectionDomain().getCodeSource().getLocation().getPath();
String decodedPath = URLDecoder.decode(path, "UTF-8");
This should solve the problem with spaces and special characters.

To obtain the File for a given Class, there are two steps:
Convert the Class to a URL
Convert the URL to a File
It is important to understand both steps, and not conflate them.
Once you have the File, you can call getParentFile to get the containing folder, if that is what you need.
Step 1: Class to URL
As discussed in other answers, there are two major ways to find a URL relevant to a Class.
URL url = Bar.class.getProtectionDomain().getCodeSource().getLocation();
URL url = Bar.class.getResource(Bar.class.getSimpleName() + ".class");
Both have pros and cons.
The getProtectionDomain approach yields the base location of the class (e.g., the containing JAR file). However, it is possible that the Java runtime's security policy will throw SecurityException when calling getProtectionDomain(), so if your application needs to run in a variety of environments, it is best to test in all of them.
The getResource approach yields the full URL resource path of the class, from which you will need to perform additional string manipulation. It may be a file: path, but it could also be jar:file: or even something nastier like bundleresource://346.fwk2106232034:4/foo/Bar.class when executing within an OSGi framework. Conversely, the getProtectionDomain approach correctly yields a file: URL even from within OSGi.
Note that both getResource("") and getResource(".") failed in my tests, when the class resided within a JAR file; both invocations returned null. So I recommend the #2 invocation shown above instead, as it seems safer.
Step 2: URL to File
Either way, once you have a URL, the next step is convert to a File. This is its own challenge; see Kohsuke Kawaguchi's blog post about it for full details, but in short, you can use new File(url.toURI()) as long as the URL is completely well-formed.
Lastly, I would highly discourage using URLDecoder. Some characters of the URL, : and / in particular, are not valid URL-encoded characters. From the URLDecoder Javadoc:
It is assumed that all characters in the encoded string are one of the following: "a" through "z", "A" through "Z", "0" through "9", and "-", "_", ".", and "*". The character "%" is allowed but is interpreted as the start of a special escaped sequence.
...
There are two possible ways in which this decoder could deal with illegal strings. It could either leave illegal characters alone or it could throw an IllegalArgumentException. Which approach the decoder takes is left to the implementation.
In practice, URLDecoder generally does not throw IllegalArgumentException as threatened above. And if your file path has spaces encoded as %20, this approach may appear to work. However, if your file path has other non-alphameric characters such as + you will have problems with URLDecoder mangling your file path.
Working code
To achieve these steps, you might have methods like the following:
/**
* Gets the base location of the given class.
* <p>
* If the class is directly on the file system (e.g.,
* "/path/to/my/package/MyClass.class") then it will return the base directory
* (e.g., "file:/path/to").
* </p>
* <p>
* If the class is within a JAR file (e.g.,
* "/path/to/my-jar.jar!/my/package/MyClass.class") then it will return the
* path to the JAR (e.g., "file:/path/to/my-jar.jar").
* </p>
*
* #param c The class whose location is desired.
* #see FileUtils#urlToFile(URL) to convert the result to a {#link File}.
*/
public static URL getLocation(final Class<?> c) {
if (c == null) return null; // could not load the class
// try the easy way first
try {
final URL codeSourceLocation =
c.getProtectionDomain().getCodeSource().getLocation();
if (codeSourceLocation != null) return codeSourceLocation;
}
catch (final SecurityException e) {
// NB: Cannot access protection domain.
}
catch (final NullPointerException e) {
// NB: Protection domain or code source is null.
}
// NB: The easy way failed, so we try the hard way. We ask for the class
// itself as a resource, then strip the class's path from the URL string,
// leaving the base path.
// get the class's raw resource path
final URL classResource = c.getResource(c.getSimpleName() + ".class");
if (classResource == null) return null; // cannot find class resource
final String url = classResource.toString();
final String suffix = c.getCanonicalName().replace('.', '/') + ".class";
if (!url.endsWith(suffix)) return null; // weird URL
// strip the class's path from the URL string
final String base = url.substring(0, url.length() - suffix.length());
String path = base;
// remove the "jar:" prefix and "!/" suffix, if present
if (path.startsWith("jar:")) path = path.substring(4, path.length() - 2);
try {
return new URL(path);
}
catch (final MalformedURLException e) {
e.printStackTrace();
return null;
}
}
/**
* Converts the given {#link URL} to its corresponding {#link File}.
* <p>
* This method is similar to calling {#code new File(url.toURI())} except that
* it also handles "jar:file:" URLs, returning the path to the JAR file.
* </p>
*
* #param url The URL to convert.
* #return A file path suitable for use with e.g. {#link FileInputStream}
* #throws IllegalArgumentException if the URL does not correspond to a file.
*/
public static File urlToFile(final URL url) {
return url == null ? null : urlToFile(url.toString());
}
/**
* Converts the given URL string to its corresponding {#link File}.
*
* #param url The URL to convert.
* #return A file path suitable for use with e.g. {#link FileInputStream}
* #throws IllegalArgumentException if the URL does not correspond to a file.
*/
public static File urlToFile(final String url) {
String path = url;
if (path.startsWith("jar:")) {
// remove "jar:" prefix and "!/" suffix
final int index = path.indexOf("!/");
path = path.substring(4, index);
}
try {
if (PlatformUtils.isWindows() && path.matches("file:[A-Za-z]:.*")) {
path = "file:/" + path.substring(5);
}
return new File(new URL(path).toURI());
}
catch (final MalformedURLException e) {
// NB: URL is not completely well-formed.
}
catch (final URISyntaxException e) {
// NB: URL is not completely well-formed.
}
if (path.startsWith("file:")) {
// pass through the URL as-is, minus "file:" prefix
path = path.substring(5);
return new File(path);
}
throw new IllegalArgumentException("Invalid URL: " + url);
}
You can find these methods in the SciJava Common library:
org.scijava.util.ClassUtils
org.scijava.util.FileUtils.

You can also use:
CodeSource codeSource = YourMainClass.class.getProtectionDomain().getCodeSource();
File jarFile = new File(codeSource.getLocation().toURI().getPath());
String jarDir = jarFile.getParentFile().getPath();

Use ClassLoader.getResource() to find the URL for your current class.
For example:
package foo;
public class Test
{
public static void main(String[] args)
{
ClassLoader loader = Test.class.getClassLoader();
System.out.println(loader.getResource("foo/Test.class"));
}
}
(This example taken from a similar question.)
To find the directory, you'd then need to take apart the URL manually. See the JarClassLoader tutorial for the format of a jar URL.

I'm surprised to see that none recently proposed to use Path. Here follows a citation: "The Path class includes various methods that can be used to obtain information about the path, access elements of the path, convert the path to other forms, or extract portions of a path"
Thus, a good alternative is to get the Path objest as:
Path path = Paths.get(Test.class.getProtectionDomain().getCodeSource().getLocation().toURI());

The only solution that works for me on Linux, Mac and Windows:
public static String getJarContainingFolder(Class aclass) throws Exception {
CodeSource codeSource = aclass.getProtectionDomain().getCodeSource();
File jarFile;
if (codeSource.getLocation() != null) {
jarFile = new File(codeSource.getLocation().toURI());
}
else {
String path = aclass.getResource(aclass.getSimpleName() + ".class").getPath();
String jarFilePath = path.substring(path.indexOf(":") + 1, path.indexOf("!"));
jarFilePath = URLDecoder.decode(jarFilePath, "UTF-8");
jarFile = new File(jarFilePath);
}
return jarFile.getParentFile().getAbsolutePath();
}

If you are really looking for a simple way to get the folder in which your JAR is located you should use this implementation.
Solutions like this are hard to find and many solutions are no longer supported, many others provide the path of the file instead of the actual directory. This is easier than other solutions you are going to find and works for java version 1.12.
new File(".").getCanonicalPath()
Gathering the Input from other answers this is a simple one too:
String localPath=new File(getClass().getProtectionDomain().getCodeSource().getLocation().toURI()).getParentFile().getPath()+"\\";
Both will return a String with this format:
"C:\Users\User\Desktop\Folder\"
In a simple and concise line.

I had the the same problem and I solved it that way:
File currentJavaJarFile = new File(Main.class.getProtectionDomain().getCodeSource().getLocation().getPath());
String currentJavaJarFilePath = currentJavaJarFile.getAbsolutePath();
String currentRootDirectoryPath = currentJavaJarFilePath.replace(currentJavaJarFile.getName(), "");
I hope I was of help to you.

Here's upgrade to other comments, that seem to me incomplete for the specifics of
using a relative "folder" outside .jar file (in the jar's same
location):
String path =
YourMainClassName.class.getProtectionDomain().
getCodeSource().getLocation().getPath();
path =
URLDecoder.decode(
path,
"UTF-8");
BufferedImage img =
ImageIO.read(
new File((
new File(path).getParentFile().getPath()) +
File.separator +
"folder" +
File.separator +
"yourfile.jpg"));

For getting the path of running jar file I have studied the above solutions and tried all methods which exist some difference each other. If these code are running in Eclipse IDE they all should be able to find the path of the file including the indicated class and open or create an indicated file with the found path.
But it is tricky, when run the runnable jar file directly or through the command line, it will be failed as the path of jar file gotten from the above methods will give an internal path in the jar file, that is it always gives a path as
rsrc:project-name (maybe I should say that it is the package name of the main class file - the indicated class)
I can not convert the rsrc:... path to an external path, that is when run the jar file outside the Eclipse IDE it can not get the path of jar file.
The only possible way for getting the path of running jar file outside Eclipse IDE is
System.getProperty("java.class.path")
this code line may return the living path (including the file name) of the running jar file (note that the return path is not the working directory), as the java document and some people said that it will return the paths of all class files in the same directory, but as my tests if in the same directory include many jar files, it only return the path of running jar (about the multiple paths issue indeed it happened in the Eclipse).

Other answers seem to point to the code source which is Jar file location which is not a directory.
Use
return new File(MyClass.class.getProtectionDomain().getCodeSource().getLocation().toURI().getPath()).getParentFile();

the selected answer above is not working if you run your jar by click on it from Gnome desktop environment (not from any script or terminal).
Instead, I have fond that the following solution is working everywhere:
try {
return URLDecoder.decode(ClassLoader.getSystemClassLoader().getResource(".").getPath(), "UTF-8");
} catch (UnsupportedEncodingException e) {
return "";
}

I had to mess around a lot before I finally found a working (and short) solution.
It is possible that the jarLocation comes with a prefix like file:\ or jar:file\, which can be removed by using String#substring().
URL jarLocationUrl = MyClass.class.getProtectionDomain().getCodeSource().getLocation();
String jarLocation = new File(jarLocationUrl.toString()).getParent();

For the jar file path:
String jarPath = new File(MyClass.class.getProtectionDomain().getCodeSource().getLocation()
.toURI()).getPath();
For getting the directory path of that jar file:
String dirPath = new File(MyClass.class.getProtectionDomain().getCodeSource().getLocation()
.toURI()).getParent();
The results of the two lines above are like this:
/home/user/MyPrograms/myapp/myjar.jar (value of jarPath)
/home/user/MyPrograms/myapp (value of dirPath)

public static String dir() throws URISyntaxException
{
URI path=Main.class.getProtectionDomain().getCodeSource().getLocation().toURI();
String name= Main.class.getPackage().getName()+".jar";
String path2 = path.getRawPath();
path2=path2.substring(1);
if (path2.contains(".jar"))
{
path2=path2.replace(name, "");
}
return path2;}
Works good on Windows

I tried to get the jar running path using
String folder = MyClassName.class.getProtectionDomain().getCodeSource().getLocation().getPath();
c:\app>java -jar application.jar
Running the jar application named "application.jar", on Windows in the folder "c:\app", the value of the String variable "folder" was "\c:\app\application.jar" and I had problems testing for path's correctness
File test = new File(folder);
if(file.isDirectory() && file.canRead()) { //always false }
So I tried to define "test" as:
String fold= new File(folder).getParentFile().getPath()
File test = new File(fold);
to get path in a right format like "c:\app" instead of "\c:\app\application.jar" and I noticed that it work.

The simplest solution is to pass the path as an argument when running the jar.
You can automate this with a shell script (.bat in Windows, .sh anywhere else):
java -jar my-jar.jar .
I used . to pass the current working directory.
UPDATE
You may want to stick the jar file in a sub-directory so users don't accidentally click it. Your code should also check to make sure that the command line arguments have been supplied, and provide a good error message if the arguments are missing.

Actually here is a better version - the old one failed if a folder name had a space in it.
private String getJarFolder() {
// get name and path
String name = getClass().getName().replace('.', '/');
name = getClass().getResource("/" + name + ".class").toString();
// remove junk
name = name.substring(0, name.indexOf(".jar"));
name = name.substring(name.lastIndexOf(':')-1, name.lastIndexOf('/')+1).replace('%', ' ');
// remove escape characters
String s = "";
for (int k=0; k<name.length(); k++) {
s += name.charAt(k);
if (name.charAt(k) == ' ') k += 2;
}
// replace '/' with system separator char
return s.replace('/', File.separatorChar);
}
As for failing with applets, you wouldn't usually have access to local files anyway. I don't know much about JWS but to handle local files might it not be possible to download the app.?

String path = getClass().getResource("").getPath();
The path always refers to the resource within the jar file.

Try this:
String path = new File("").getAbsolutePath();

This code worked for me to identify if the program is being executed inside a JAR file or IDE:
private static boolean isRunningOverJar() {
try {
String pathJar = Application.class.getResource(Application.class.getSimpleName() + ".class").getFile();
if (pathJar.toLowerCase().contains(".jar")) {
return true;
} else {
return false;
}
} catch (Exception e) {
return false;
}
}
If I need to get the Windows full path of JAR file I am using this method:
private static String getPathJar() {
try {
final URI jarUriPath =
Application.class.getResource(Application.class.getSimpleName() + ".class").toURI();
String jarStringPath = jarUriPath.toString().replace("jar:", "");
String jarCleanPath = Paths.get(new URI(jarStringPath)).toString();
if (jarCleanPath.toLowerCase().contains(".jar")) {
return jarCleanPath.substring(0, jarCleanPath.lastIndexOf(".jar") + 4);
} else {
return null;
}
} catch (Exception e) {
log.error("Error getting JAR path.", e);
return null;
}
}
My complete code working with a Spring Boot application using CommandLineRunner implementation, to ensure that the application always be executed within of a console view (Double clicks by mistake in JAR file name), I am using the next code:
#SpringBootApplication
public class Application implements CommandLineRunner {
public static void main(String[] args) throws IOException {
Console console = System.console();
if (console == null && !GraphicsEnvironment.isHeadless() && isRunningOverJar()) {
Runtime.getRuntime().exec(new String[]{"cmd", "/c", "start", "cmd", "/k",
"java -jar \"" + getPathJar() + "\""});
} else {
SpringApplication.run(Application.class, args);
}
}
#Override
public void run(String... args) {
/*
Additional code here...
*/
}
private static boolean isRunningOverJar() {
try {
String pathJar = Application.class.getResource(Application.class.getSimpleName() + ".class").getFile();
if (pathJar.toLowerCase().contains(".jar")) {
return true;
} else {
return false;
}
} catch (Exception e) {
return false;
}
}
private static String getPathJar() {
try {
final URI jarUriPath =
Application.class.getResource(Application.class.getSimpleName() + ".class").toURI();
String jarStringPath = jarUriPath.toString().replace("jar:", "");
String jarCleanPath = Paths.get(new URI(jarStringPath)).toString();
if (jarCleanPath.toLowerCase().contains(".jar")) {
return jarCleanPath.substring(0, jarCleanPath.lastIndexOf(".jar") + 4);
} else {
return null;
}
} catch (Exception e) {
return null;
}
}
}

Something that is frustrating is that when you are developing in Eclipse MyClass.class.getProtectionDomain().getCodeSource().getLocation() returns the /bin directory which is great, but when you compile it to a jar, the path includes the /myjarname.jar part which gives you illegal file names.
To have the code work both in the ide and once it is compiled to a jar, I use the following piece of code:
URL applicationRootPathURL = getClass().getProtectionDomain().getCodeSource().getLocation();
File applicationRootPath = new File(applicationRootPathURL.getPath());
File myFile;
if(applicationRootPath.isDirectory()){
myFile = new File(applicationRootPath, "filename");
}
else{
myFile = new File(applicationRootPath.getParentFile(), "filename");
}

Not really sure about the others but in my case it didn't work with a "Runnable jar" and i got it working by fixing codes together from phchen2 answer and another from this link :How to get the path of a running JAR file?
The code:
String path=new java.io.File(Server.class.getProtectionDomain()
.getCodeSource()
.getLocation()
.getPath())
.getAbsolutePath();
path=path.substring(0, path.lastIndexOf("."));
path=path+System.getProperty("java.class.path");

Have tried several of the solutions up there but none yielded correct results for the (probably special) case that the runnable jar has been exported with "Packaging external libraries" in Eclipse. For some reason all solutions based on the ProtectionDomain do result in null in that case.
From combining some solutions above I managed to achieve the following working code:
String surroundingJar = null;
// gets the path to the jar file if it exists; or the "bin" directory if calling from Eclipse
String jarDir = new File(ClassLoader.getSystemClassLoader().getResource(".").getPath()).getAbsolutePath();
// gets the "bin" directory if calling from eclipse or the name of the .jar file alone (without its path)
String jarFileFromSys = System.getProperty("java.class.path").split(";")[0];
// If both are equal that means it is running from an IDE like Eclipse
if (jarFileFromSys.equals(jarDir))
{
System.out.println("RUNNING FROM IDE!");
// The path to the jar is the "bin" directory in that case because there is no actual .jar file.
surroundingJar = jarDir;
}
else
{
// Combining the path and the name of the .jar file to achieve the final result
surroundingJar = jarDir + jarFileFromSys.substring(1);
}
System.out.println("JAR File: " + surroundingJar);

The above methods didn't work for me in my Spring environment, since Spring shades the actual classes into a package called BOOT-INF, thus not the actual location of the running file. I found another way to retrieve the running file through the Permissions object which have been granted to the running file:
public static Path getEnclosingDirectory() {
return Paths.get(FileUtils.class.getProtectionDomain().getPermissions()
.elements().nextElement().getName()).getParent();
}

Mention that it is checked only in Windows but i think it works perfect on other Operating Systems [Linux,MacOs,Solaris] :).
I had 2 .jar files in the same directory . I wanted from the one .jar file to start the other .jar file which is in the same directory.
The problem is that when you start it from the cmd the current directory is system32.
Warnings!
The below seems to work pretty well in all the test i have done even
with folder name ;][[;'57f2g34g87-8+9-09!2##!$%^^&() or ()%&$%^##
it works well.
I am using the ProcessBuilder with the below as following:
🍂..
//The class from which i called this was the class `Main`
String path = getBasePathForClass(Main.class);
String applicationPath= new File(path + "application.jar").getAbsolutePath();
System.out.println("Directory Path is : "+applicationPath);
//Your know try catch here
//Mention that sometimes it doesn't work for example with folder `;][[;'57f2g34g87-8+9-09!2##!$%^^&()`
ProcessBuilder builder = new ProcessBuilder("java", "-jar", applicationPath);
builder.redirectErrorStream(true);
Process process = builder.start();
//...code
🍂getBasePathForClass(Class<?> classs):
/**
* Returns the absolute path of the current directory in which the given
* class
* file is.
*
* #param classs
* #return The absolute path of the current directory in which the class
* file is.
* #author GOXR3PLUS[StackOverFlow user] + bachden [StackOverFlow user]
*/
public static final String getBasePathForClass(Class<?> classs) {
// Local variables
File file;
String basePath = "";
boolean failed = false;
// Let's give a first try
try {
file = new File(classs.getProtectionDomain().getCodeSource().getLocation().toURI().getPath());
if (file.isFile() || file.getPath().endsWith(".jar") || file.getPath().endsWith(".zip")) {
basePath = file.getParent();
} else {
basePath = file.getPath();
}
} catch (URISyntaxException ex) {
failed = true;
Logger.getLogger(classs.getName()).log(Level.WARNING,
"Cannot firgue out base path for class with way (1): ", ex);
}
// The above failed?
if (failed) {
try {
file = new File(classs.getClassLoader().getResource("").toURI().getPath());
basePath = file.getAbsolutePath();
// the below is for testing purposes...
// starts with File.separator?
// String l = local.replaceFirst("[" + File.separator +
// "/\\\\]", "")
} catch (URISyntaxException ex) {
Logger.getLogger(classs.getName()).log(Level.WARNING,
"Cannot firgue out base path for class with way (2): ", ex);
}
}
// fix to run inside eclipse
if (basePath.endsWith(File.separator + "lib") || basePath.endsWith(File.separator + "bin")
|| basePath.endsWith("bin" + File.separator) || basePath.endsWith("lib" + File.separator)) {
basePath = basePath.substring(0, basePath.length() - 4);
}
// fix to run inside netbeans
if (basePath.endsWith(File.separator + "build" + File.separator + "classes")) {
basePath = basePath.substring(0, basePath.length() - 14);
}
// end fix
if (!basePath.endsWith(File.separator)) {
basePath = basePath + File.separator;
}
return basePath;
}

This code worked for me:
private static String getJarPath() throws IOException, URISyntaxException {
File f = new File(LicensingApp.class.getProtectionDomain().().getLocation().toURI());
String jarPath = f.getCanonicalPath().toString();
String jarDir = jarPath.substring( 0, jarPath.lastIndexOf( File.separator ));
return jarDir;
}

The getProtectionDomain approach might not work sometimes e.g. when you have to find the jar for some of the core java classes (e.g in my case StringBuilder class within IBM JDK), however following works seamlessly:
public static void main(String[] args) {
System.out.println(findSource(MyClass.class));
// OR
System.out.println(findSource(String.class));
}
public static String findSource(Class<?> clazz) {
String resourceToSearch = '/' + clazz.getName().replace(".", "/") + ".class";
java.net.URL location = clazz.getResource(resourceToSearch);
String sourcePath = location.getPath();
// Optional, Remove junk
return sourcePath.replace("file:", "").replace("!" + resourceToSearch, "");
}

I have another way to get the String location of a class.
URL path = Thread.currentThread().getContextClassLoader().getResource("");
Path p = Paths.get(path.toURI());
String location = p.toString();
The output String will have the form of
C:\Users\Administrator\new Workspace\...
The spaces and other characters are handled, and in the form without file:/. So will be easier to use.

Processing.org 3 and loading Java properties file?

I am not all that knowledgeable in Java, and I'm trying to use some Java in a Processing.org 3 project. I have managed to reconstruct the problem in a small Processing example project called testprocjavapath - and I am posting a bash script (called testProcJavaLoadpath.sh) that reconstructs the example project files at the end of this post, and runs the project once. The testprocjavapath project files look like this:
~/sketchbook/testprocjavapath
├── testprocjavapath.pde
├── myprops.properties
└── MyJavaClass.java
When running the script, I get this:
$ bash testProcJavaLoadpath.sh
...
There was an exception myprops.properties: java.lang.NullPointerException : null
The properties file content is 'null';
Finished.
Debugging in Processing 3 IDE GUI, this error occurs on exactly the line properties.load(in);:
... because the line InputStream in = MyJavaClass.class.getResourceAsStream(inFileName); failed, and as a result, in is a null pointer.
That much I understand - what I don't understand is this: how do I load a, say, .properties text file, in the same directory as the .pde Processing sketch file and the .java file (that is, this particular sketch folder)?
As far as I gather the Java getResourceAsStream() actually is used to load from a Java application packed as a .jar file - so can it work for reading files from hard disk, that are not yet packed as .jar files?
If not - I have also tried to do:
InputStream in = new FileInputStream( new File(PROPERTIES_FILENAME));
... but this didn't work either (in is again null).
So, what command can I use in the .java file, to load the myprops.properties file? And if I should end up packing the whole Processing app as a .jar file (not sure if Processing can do this, haven't looked it up yet), would I have to change that command?
Here is the testProcJavaLoadpath.sh file (make sure you change PROCBINPATH to your Processing install path):
PROCSKETCHDIR="~/sketchbook"
PROCSKETCHDIR="${PROCSKETCHDIR/#\~/$HOME}" # expand home dir ~
echo "$PROCSKETCHDIR"
PROCBINPATH="/PATH/TO/processing-3.3.6" # path/location of Processing executable `processing-java`
MYSKETCH="testprocjavapath"
MYSKETCHDIR="$PROCSKETCHDIR/$MYSKETCH"
# reconstruct folder:
rm -rfv "$MYSKETCHDIR"
mkdir -v "$MYSKETCHDIR"
echo "generating $MYSKETCHDIR/$MYSKETCH.pde"
cat > "$MYSKETCHDIR/$MYSKETCH.pde" <<'EOF'
void setup() {
size(640, 360); // Size should be the first statement
MyJavaClass myjc = new MyJavaClass();
String thefilecontents = myjc.GetPropsFileContent();
System.out.format("The properties file content is '%s';%n", thefilecontents);
}
EOF
echo "generating $MYSKETCHDIR/myprops.properties"
cat > "$MYSKETCHDIR/myprops.properties" <<'EOF'
teststr=HelloWorld
EOF
echo "generating $MYSKETCHDIR/MyJavaClass.java"
cat > "$MYSKETCHDIR/MyJavaClass.java" <<'EOF'
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.Iterator;
import java.util.List;
import java.util.Properties;
import java.io.FileInputStream;
import java.io.File;
import java.io.ByteArrayOutputStream;
public class MyJavaClass {
private static final String PROPERTIES_FILENAME = "myprops.properties";
/**
* add a constructor
*/
public static void MyJavaClass() {
}
public static String GetPropsFileContent() {
String myret = null;
myret = readgetFileContent(PROPERTIES_FILENAME);
return myret;
}
public static String readgetFileContent(String inFileName) {
String result = null;
Properties properties = new Properties();
try {
InputStream in = MyJavaClass.class.getResourceAsStream(inFileName);
properties.load(in);
ByteArrayOutputStream resultbaos = new ByteArrayOutputStream();
byte[] buffer = new byte[1024];
int length;
while ((length = in.read(buffer)) != -1) {
resultbaos.write(buffer, 0, length);
}
result = resultbaos.toString();
} catch (IOException e) {
System.err.println("There was an error reading " + inFileName + ": " + e.getCause()
+ " : " + e.getMessage());
} catch (Exception e) {
System.err.println("There was an exception " + inFileName + ": " + e
+ " : " + e.getMessage());
}
return result;
}
}
EOF
# run once:
"$PROCBINPATH"/processing-java --sketch="$MYSKETCHDIR" --run

Ok, I managed to get somewhere - but I'd still like a more qualified answer.
First of all, it turns out, Processing as such expect files to be read (like myprops.properties in the OP example), to be stored in a data subfolder of the sketch folder:
https://processing.org/tutorials/data/
And just as with image files, these text files should be placed in the sketch’s “data” directory in order for them to be recognized by the Processing sketch.
So far so good - and indeed, inside the Processing .pde sketch, we can use (say) loadStrings("myprops.properties");, and the file at data/myprops.properties will be read. However, I don't need to read the file there - I need to read it in the supporting .java class.
Now, when you run the Processing patch (either from IDE, or from the command line), what happens is that Processing copies the source files from the sketch folder, inside a temporary folder in the /tmp folder (at least on Linux); here's how that file structure looks like:
/tmp/testprocjavapath9179591342074530534temp/
├── MyJavaClass.class
├── source
│   ├── MyJavaClass.java
│   └── testprocjavapath.java
└── testprocjavapath.class
Notice that we have .java source files, and .class "compiled" files, but there is no data subfolder or myprops.properties file anywhere!
Now, notice also that what used to be testprocjavapath.pde in the source sketch folder, becomes testprocjavapath.java (and corresponding .class) in the temporary folder; notice that testprocjavapath.java defines:
public class testprocjavapath extends PApplet {
Now, the loadStrings is actually a method of the PApplet class; so if we read through it a bit:
https://github.com/processing/processing/blob/master/core/src/processing/core/PApplet.java
dataPath(String where): .... The data path is handled differently on each platform, and should not be considered a location to write files. It should also not be assumed that this location can be read from or listed. ... Libraries should use createInput() to get an InputStream or createOutput() to get an OutputStream. sketchPath() can be used to get a location relative to the sketch. Again, do not use this to get relative locations of files. ...
... we can see a method dataPath, however it's usage is not recommended. On the other hand, there is a method sketchPath - however, this method will only return the correct path (i.e. a sketch in ~/sketchbook in this example) if called from the top-level .pde file! If you try to have the class in the .java file defined as extends PApplet, and then call sketchPath from there - it will simply return the current working directory!
So the solution for now is:
Have the .java class accept an input argument in the constructor, which will be used to record the proper sketch path: public MyJavaClass(String inSketchPath) {
Then, have the sketchPath() passed to the .java class instance during instantiation in the .pde file: MyJavaClass myjc = new MyJavaClass( sketchPath() );
Finally, use the passed sketch path inside the .java class, to calculate the absolute path of the .properties file, and then load it with new FileInputStream( new File(theFilePath) ); (not with getResourceAsStream!)
Below a changed testProcJavaLoadpath.sh is pasted, that has these modifications, and in principle, works - this is the terminal output:
$ bash testProcJavaLoadpath.sh
...
Sketch first lines: 'teststr=HelloWorld';
Sketch dataFile: '~/sketchbook/testprocjavapath/data/myprops.properties';
Sketch sketchPath: '~/sketchbook/testprocjavapath';
:: mySketchPath: '~/sketchbook/testprocjavapath'
:: The URL is 'file:/tmp/testprocjavapath4709659129218148940temp/';
:: name: MyJavaClass.class
:: resourcePath: file:/tmp/testprocjavapath4709659129218148940temp/MyJavaClass.class
:: theFilePath: '~/sketchbook/testprocjavapath/data/myprops.properties'
:: properties: key 'teststr' => value 'HelloWorld'
The properties file content is 'teststr=HelloWorld
';
... however, I imagine if I should want to pack this code in a .jar or executable application/file, this approach would likely fail - which is why I'd still like a more qualified answer.
The changed testProcJavaLoadpath.sh is this:
PROCSKETCHDIR="~/sketchbook"
PROCSKETCHDIR="${PROCSKETCHDIR/#\~/$HOME}" # expand home dir ~
echo "$PROCSKETCHDIR"
PROCBINPATH="/PATH/TO/processing-3.3.6" # path/location of Processing executable `processing-java`
MYSKETCH="testprocjavapath"
MYSKETCHDIR="$PROCSKETCHDIR/$MYSKETCH"
# reconstruct folder:
rm -rfv "$MYSKETCHDIR"
mkdir -v "$MYSKETCHDIR"
# https://processing.org/tutorials/data/
# "And just as with image files, these text files should be placed in the sketch’s “data” directory in order for them to be recognized by the Processing sketch."
# processing.core.PApplet.loadStrings - https://processing.github.io/processing-javadocs/core/
# https://github.com/processing/processing/blob/master/core/src/processing/core/PApplet.java
# "dataPath(String where): The data path is handled differently on each platform, and should not be considered a location to write files. It should also not be assumed that this location can be read from or listed. ... Libraries should use createInput() to get an InputStream or createOutput() to get an OutputStream. sketchPath() can be used to get a location relative to the sketch. Again, <b>do not</b> use this to get relative locations of files."
echo "generating $MYSKETCHDIR/$MYSKETCH.pde"
cat > "$MYSKETCHDIR/$MYSKETCH.pde" <<'EOF'
void setup() {
size(640, 360); // Size should be the first statement
String[] lines = loadStrings("myprops.properties"); // reads from data/myprops.properties
System.out.format("Sketch first lines: '%s';%n", lines[0]);
System.out.format("Sketch dataFile: '%s';%n", dataFile("myprops.properties")); // ~/sketchbook/testprocjavapath/data/myprops.properties
System.out.format("Sketch sketchPath: '%s';%n", sketchPath()); // ~/sketchbook/testprocjavapath
MyJavaClass myjc = new MyJavaClass( sketchPath() );
String thefilecontents = myjc.GetPropsFileContent();
System.out.format("The properties file content is '%s';%n", thefilecontents);
}
EOF
mkdir -v "$MYSKETCHDIR/data"
echo "generating $MYSKETCHDIR/data/myprops.properties"
cat > "$MYSKETCHDIR/data/myprops.properties" <<'EOF'
teststr=HelloWorld
EOF
echo "generating $MYSKETCHDIR/MyJavaClass.java"
cat > "$MYSKETCHDIR/MyJavaClass.java" <<'EOF'
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream; // "InputStream is by definition not seekable."
import java.io.InputStreamReader;
import java.util.Iterator;
import java.util.List;
import java.util.Properties;
import java.io.FileInputStream; // is seekable
import java.io.File;
import java.io.ByteArrayOutputStream;
import java.net.URL;
//import processing.core.*;
import java.nio.file.Path;
import java.nio.file.Paths;
public class MyJavaClass {
private static final String PROPERTIES_FILENAME = "myprops.properties";
public String mySketchPath;
/**
* add a constructor
*/
public MyJavaClass(String inSketchPath) {
mySketchPath = inSketchPath;
}
public String GetPropsFileContent() {
//System.out.format(":: sketchPath: '%s'%n", sketchPath()); // if `MyJavaClass extends PApplet`, then sketchPath() just prints current working directory!
System.out.format(":: mySketchPath: '%s'%n", mySketchPath);
getLocations();
String myret = null;
myret = readgetFileContent(PROPERTIES_FILENAME);
return myret;
}
public String readgetFileContent(String inFileName) {
String result = null;
Properties properties = new Properties();
try {
//String theFilePath = inFileName; // verbatim relative path fails
Path inFileNameSketchPath = Paths.get(mySketchPath, "data", inFileName); // OS path join
String theFilePath = inFileNameSketchPath.toString();
System.out.format(":: theFilePath: '%s'%n", theFilePath);
//InputStream in = MyJavaClass.class.getResourceAsStream(theFilePath); // no can do, is 'null', also w/ abs path
//InputStream in = new FileInputStream( new File(theFilePath) ); // OK, but not seekable
FileInputStream in = new FileInputStream( new File(theFilePath) );
properties.load(in);
// double-check loaded properties:
for(String key : properties.stringPropertyNames()) {
String value = properties.getProperty(key);
System.out.format(":: properties: key '%s' => value '%s'%n", key, value);
}
ByteArrayOutputStream resultbaos = new ByteArrayOutputStream();
byte[] buffer = new byte[1024];
int length;
in.getChannel().position(0); // do reset - seek 0 (start), to reset stream again for reading
while ((length = in.read(buffer)) != -1) {
resultbaos.write(buffer, 0, length);
}
result = resultbaos.toString();
} catch (IOException e) {
System.err.println("There was an error reading " + inFileName + ": " + e.getCause()
+ " : " + e.getMessage());
} catch (Exception e) {
System.err.println("There was an exception " + inFileName + ": " + e
+ " : " + e.getMessage());
}
return result;
}
public void getLocations() {
URL classURL = getClass().getProtectionDomain().getCodeSource().getLocation();
System.out.format(":: The URL is '%s';%n", classURL); // file:/tmp/testprocjavapath3056820301028631180temp/
String s = getClass().getName();
int i = s.lastIndexOf(".");
if(i > -1) s = s.substring(i + 1);
s = s + ".class";
System.out.println(":: name: " + s);
Object resourcePath = this.getClass().getResource(s);
System.out.println(":: resourcePath: " + resourcePath); // file:/tmp/testprocjavapath9185318125154993853temp/MyJavaClass.class
}
}
EOF
# run once:
"$PROCBINPATH"/processing-java --sketch="$MYSKETCHDIR" --run

Java check if the application is getting run from a specific phat [duplicate]

Want to improve this post? Provide detailed answers to this question, including citations and an explanation of why your answer is correct. Answers without enough detail may be edited or deleted.
My code runs inside a JAR file, say foo.jar, and I need to know, in the code, in which folder the running foo.jar is.
So, if foo.jar is in C:\FOO\, I want to get that path no matter what my current working directory is.

return new File(MyClass.class.getProtectionDomain().getCodeSource().getLocation()
.toURI()).getPath();
Replace "MyClass" with the name of your class.
Obviously, this will do odd things if your class was loaded from a non-file location.

Best solution for me:
String path = Test.class.getProtectionDomain().getCodeSource().getLocation().getPath();
String decodedPath = URLDecoder.decode(path, "UTF-8");
This should solve the problem with spaces and special characters.

To obtain the File for a given Class, there are two steps:
Convert the Class to a URL
Convert the URL to a File
It is important to understand both steps, and not conflate them.
Once you have the File, you can call getParentFile to get the containing folder, if that is what you need.
Step 1: Class to URL
As discussed in other answers, there are two major ways to find a URL relevant to a Class.
URL url = Bar.class.getProtectionDomain().getCodeSource().getLocation();
URL url = Bar.class.getResource(Bar.class.getSimpleName() + ".class");
Both have pros and cons.
The getProtectionDomain approach yields the base location of the class (e.g., the containing JAR file). However, it is possible that the Java runtime's security policy will throw SecurityException when calling getProtectionDomain(), so if your application needs to run in a variety of environments, it is best to test in all of them.
The getResource approach yields the full URL resource path of the class, from which you will need to perform additional string manipulation. It may be a file: path, but it could also be jar:file: or even something nastier like bundleresource://346.fwk2106232034:4/foo/Bar.class when executing within an OSGi framework. Conversely, the getProtectionDomain approach correctly yields a file: URL even from within OSGi.
Note that both getResource("") and getResource(".") failed in my tests, when the class resided within a JAR file; both invocations returned null. So I recommend the #2 invocation shown above instead, as it seems safer.
Step 2: URL to File
Either way, once you have a URL, the next step is convert to a File. This is its own challenge; see Kohsuke Kawaguchi's blog post about it for full details, but in short, you can use new File(url.toURI()) as long as the URL is completely well-formed.
Lastly, I would highly discourage using URLDecoder. Some characters of the URL, : and / in particular, are not valid URL-encoded characters. From the URLDecoder Javadoc:
It is assumed that all characters in the encoded string are one of the following: "a" through "z", "A" through "Z", "0" through "9", and "-", "_", ".", and "*". The character "%" is allowed but is interpreted as the start of a special escaped sequence.
...
There are two possible ways in which this decoder could deal with illegal strings. It could either leave illegal characters alone or it could throw an IllegalArgumentException. Which approach the decoder takes is left to the implementation.
In practice, URLDecoder generally does not throw IllegalArgumentException as threatened above. And if your file path has spaces encoded as %20, this approach may appear to work. However, if your file path has other non-alphameric characters such as + you will have problems with URLDecoder mangling your file path.
Working code
To achieve these steps, you might have methods like the following:
/**
* Gets the base location of the given class.
* <p>
* If the class is directly on the file system (e.g.,
* "/path/to/my/package/MyClass.class") then it will return the base directory
* (e.g., "file:/path/to").
* </p>
* <p>
* If the class is within a JAR file (e.g.,
* "/path/to/my-jar.jar!/my/package/MyClass.class") then it will return the
* path to the JAR (e.g., "file:/path/to/my-jar.jar").
* </p>
*
* #param c The class whose location is desired.
* #see FileUtils#urlToFile(URL) to convert the result to a {#link File}.
*/
public static URL getLocation(final Class<?> c) {
if (c == null) return null; // could not load the class
// try the easy way first
try {
final URL codeSourceLocation =
c.getProtectionDomain().getCodeSource().getLocation();
if (codeSourceLocation != null) return codeSourceLocation;
}
catch (final SecurityException e) {
// NB: Cannot access protection domain.
}
catch (final NullPointerException e) {
// NB: Protection domain or code source is null.
}
// NB: The easy way failed, so we try the hard way. We ask for the class
// itself as a resource, then strip the class's path from the URL string,
// leaving the base path.
// get the class's raw resource path
final URL classResource = c.getResource(c.getSimpleName() + ".class");
if (classResource == null) return null; // cannot find class resource
final String url = classResource.toString();
final String suffix = c.getCanonicalName().replace('.', '/') + ".class";
if (!url.endsWith(suffix)) return null; // weird URL
// strip the class's path from the URL string
final String base = url.substring(0, url.length() - suffix.length());
String path = base;
// remove the "jar:" prefix and "!/" suffix, if present
if (path.startsWith("jar:")) path = path.substring(4, path.length() - 2);
try {
return new URL(path);
}
catch (final MalformedURLException e) {
e.printStackTrace();
return null;
}
}
/**
* Converts the given {#link URL} to its corresponding {#link File}.
* <p>
* This method is similar to calling {#code new File(url.toURI())} except that
* it also handles "jar:file:" URLs, returning the path to the JAR file.
* </p>
*
* #param url The URL to convert.
* #return A file path suitable for use with e.g. {#link FileInputStream}
* #throws IllegalArgumentException if the URL does not correspond to a file.
*/
public static File urlToFile(final URL url) {
return url == null ? null : urlToFile(url.toString());
}
/**
* Converts the given URL string to its corresponding {#link File}.
*
* #param url The URL to convert.
* #return A file path suitable for use with e.g. {#link FileInputStream}
* #throws IllegalArgumentException if the URL does not correspond to a file.
*/
public static File urlToFile(final String url) {
String path = url;
if (path.startsWith("jar:")) {
// remove "jar:" prefix and "!/" suffix
final int index = path.indexOf("!/");
path = path.substring(4, index);
}
try {
if (PlatformUtils.isWindows() && path.matches("file:[A-Za-z]:.*")) {
path = "file:/" + path.substring(5);
}
return new File(new URL(path).toURI());
}
catch (final MalformedURLException e) {
// NB: URL is not completely well-formed.
}
catch (final URISyntaxException e) {
// NB: URL is not completely well-formed.
}
if (path.startsWith("file:")) {
// pass through the URL as-is, minus "file:" prefix
path = path.substring(5);
return new File(path);
}
throw new IllegalArgumentException("Invalid URL: " + url);
}
You can find these methods in the SciJava Common library:
org.scijava.util.ClassUtils
org.scijava.util.FileUtils.

You can also use:
CodeSource codeSource = YourMainClass.class.getProtectionDomain().getCodeSource();
File jarFile = new File(codeSource.getLocation().toURI().getPath());
String jarDir = jarFile.getParentFile().getPath();

Use ClassLoader.getResource() to find the URL for your current class.
For example:
package foo;
public class Test
{
public static void main(String[] args)
{
ClassLoader loader = Test.class.getClassLoader();
System.out.println(loader.getResource("foo/Test.class"));
}
}
(This example taken from a similar question.)
To find the directory, you'd then need to take apart the URL manually. See the JarClassLoader tutorial for the format of a jar URL.

I'm surprised to see that none recently proposed to use Path. Here follows a citation: "The Path class includes various methods that can be used to obtain information about the path, access elements of the path, convert the path to other forms, or extract portions of a path"
Thus, a good alternative is to get the Path objest as:
Path path = Paths.get(Test.class.getProtectionDomain().getCodeSource().getLocation().toURI());

The only solution that works for me on Linux, Mac and Windows:
public static String getJarContainingFolder(Class aclass) throws Exception {
CodeSource codeSource = aclass.getProtectionDomain().getCodeSource();
File jarFile;
if (codeSource.getLocation() != null) {
jarFile = new File(codeSource.getLocation().toURI());
}
else {
String path = aclass.getResource(aclass.getSimpleName() + ".class").getPath();
String jarFilePath = path.substring(path.indexOf(":") + 1, path.indexOf("!"));
jarFilePath = URLDecoder.decode(jarFilePath, "UTF-8");
jarFile = new File(jarFilePath);
}
return jarFile.getParentFile().getAbsolutePath();
}

If you are really looking for a simple way to get the folder in which your JAR is located you should use this implementation.
Solutions like this are hard to find and many solutions are no longer supported, many others provide the path of the file instead of the actual directory. This is easier than other solutions you are going to find and works for java version 1.12.
new File(".").getCanonicalPath()
Gathering the Input from other answers this is a simple one too:
String localPath=new File(getClass().getProtectionDomain().getCodeSource().getLocation().toURI()).getParentFile().getPath()+"\\";
Both will return a String with this format:
"C:\Users\User\Desktop\Folder\"
In a simple and concise line.

I had the the same problem and I solved it that way:
File currentJavaJarFile = new File(Main.class.getProtectionDomain().getCodeSource().getLocation().getPath());
String currentJavaJarFilePath = currentJavaJarFile.getAbsolutePath();
String currentRootDirectoryPath = currentJavaJarFilePath.replace(currentJavaJarFile.getName(), "");
I hope I was of help to you.

Here's upgrade to other comments, that seem to me incomplete for the specifics of
using a relative "folder" outside .jar file (in the jar's same
location):
String path =
YourMainClassName.class.getProtectionDomain().
getCodeSource().getLocation().getPath();
path =
URLDecoder.decode(
path,
"UTF-8");
BufferedImage img =
ImageIO.read(
new File((
new File(path).getParentFile().getPath()) +
File.separator +
"folder" +
File.separator +
"yourfile.jpg"));

For getting the path of running jar file I have studied the above solutions and tried all methods which exist some difference each other. If these code are running in Eclipse IDE they all should be able to find the path of the file including the indicated class and open or create an indicated file with the found path.
But it is tricky, when run the runnable jar file directly or through the command line, it will be failed as the path of jar file gotten from the above methods will give an internal path in the jar file, that is it always gives a path as
rsrc:project-name (maybe I should say that it is the package name of the main class file - the indicated class)
I can not convert the rsrc:... path to an external path, that is when run the jar file outside the Eclipse IDE it can not get the path of jar file.
The only possible way for getting the path of running jar file outside Eclipse IDE is
System.getProperty("java.class.path")
this code line may return the living path (including the file name) of the running jar file (note that the return path is not the working directory), as the java document and some people said that it will return the paths of all class files in the same directory, but as my tests if in the same directory include many jar files, it only return the path of running jar (about the multiple paths issue indeed it happened in the Eclipse).

Other answers seem to point to the code source which is Jar file location which is not a directory.
Use
return new File(MyClass.class.getProtectionDomain().getCodeSource().getLocation().toURI().getPath()).getParentFile();

the selected answer above is not working if you run your jar by click on it from Gnome desktop environment (not from any script or terminal).
Instead, I have fond that the following solution is working everywhere:
try {
return URLDecoder.decode(ClassLoader.getSystemClassLoader().getResource(".").getPath(), "UTF-8");
} catch (UnsupportedEncodingException e) {
return "";
}

I had to mess around a lot before I finally found a working (and short) solution.
It is possible that the jarLocation comes with a prefix like file:\ or jar:file\, which can be removed by using String#substring().
URL jarLocationUrl = MyClass.class.getProtectionDomain().getCodeSource().getLocation();
String jarLocation = new File(jarLocationUrl.toString()).getParent();

For the jar file path:
String jarPath = new File(MyClass.class.getProtectionDomain().getCodeSource().getLocation()
.toURI()).getPath();
For getting the directory path of that jar file:
String dirPath = new File(MyClass.class.getProtectionDomain().getCodeSource().getLocation()
.toURI()).getParent();
The results of the two lines above are like this:
/home/user/MyPrograms/myapp/myjar.jar (value of jarPath)
/home/user/MyPrograms/myapp (value of dirPath)

public static String dir() throws URISyntaxException
{
URI path=Main.class.getProtectionDomain().getCodeSource().getLocation().toURI();
String name= Main.class.getPackage().getName()+".jar";
String path2 = path.getRawPath();
path2=path2.substring(1);
if (path2.contains(".jar"))
{
path2=path2.replace(name, "");
}
return path2;}
Works good on Windows

I tried to get the jar running path using
String folder = MyClassName.class.getProtectionDomain().getCodeSource().getLocation().getPath();
c:\app>java -jar application.jar
Running the jar application named "application.jar", on Windows in the folder "c:\app", the value of the String variable "folder" was "\c:\app\application.jar" and I had problems testing for path's correctness
File test = new File(folder);
if(file.isDirectory() && file.canRead()) { //always false }
So I tried to define "test" as:
String fold= new File(folder).getParentFile().getPath()
File test = new File(fold);
to get path in a right format like "c:\app" instead of "\c:\app\application.jar" and I noticed that it work.

The simplest solution is to pass the path as an argument when running the jar.
You can automate this with a shell script (.bat in Windows, .sh anywhere else):
java -jar my-jar.jar .
I used . to pass the current working directory.
UPDATE
You may want to stick the jar file in a sub-directory so users don't accidentally click it. Your code should also check to make sure that the command line arguments have been supplied, and provide a good error message if the arguments are missing.

Actually here is a better version - the old one failed if a folder name had a space in it.
private String getJarFolder() {
// get name and path
String name = getClass().getName().replace('.', '/');
name = getClass().getResource("/" + name + ".class").toString();
// remove junk
name = name.substring(0, name.indexOf(".jar"));
name = name.substring(name.lastIndexOf(':')-1, name.lastIndexOf('/')+1).replace('%', ' ');
// remove escape characters
String s = "";
for (int k=0; k<name.length(); k++) {
s += name.charAt(k);
if (name.charAt(k) == ' ') k += 2;
}
// replace '/' with system separator char
return s.replace('/', File.separatorChar);
}
As for failing with applets, you wouldn't usually have access to local files anyway. I don't know much about JWS but to handle local files might it not be possible to download the app.?

String path = getClass().getResource("").getPath();
The path always refers to the resource within the jar file.

Try this:
String path = new File("").getAbsolutePath();

This code worked for me to identify if the program is being executed inside a JAR file or IDE:
private static boolean isRunningOverJar() {
try {
String pathJar = Application.class.getResource(Application.class.getSimpleName() + ".class").getFile();
if (pathJar.toLowerCase().contains(".jar")) {
return true;
} else {
return false;
}
} catch (Exception e) {
return false;
}
}
If I need to get the Windows full path of JAR file I am using this method:
private static String getPathJar() {
try {
final URI jarUriPath =
Application.class.getResource(Application.class.getSimpleName() + ".class").toURI();
String jarStringPath = jarUriPath.toString().replace("jar:", "");
String jarCleanPath = Paths.get(new URI(jarStringPath)).toString();
if (jarCleanPath.toLowerCase().contains(".jar")) {
return jarCleanPath.substring(0, jarCleanPath.lastIndexOf(".jar") + 4);
} else {
return null;
}
} catch (Exception e) {
log.error("Error getting JAR path.", e);
return null;
}
}
My complete code working with a Spring Boot application using CommandLineRunner implementation, to ensure that the application always be executed within of a console view (Double clicks by mistake in JAR file name), I am using the next code:
#SpringBootApplication
public class Application implements CommandLineRunner {
public static void main(String[] args) throws IOException {
Console console = System.console();
if (console == null && !GraphicsEnvironment.isHeadless() && isRunningOverJar()) {
Runtime.getRuntime().exec(new String[]{"cmd", "/c", "start", "cmd", "/k",
"java -jar \"" + getPathJar() + "\""});
} else {
SpringApplication.run(Application.class, args);
}
}
#Override
public void run(String... args) {
/*
Additional code here...
*/
}
private static boolean isRunningOverJar() {
try {
String pathJar = Application.class.getResource(Application.class.getSimpleName() + ".class").getFile();
if (pathJar.toLowerCase().contains(".jar")) {
return true;
} else {
return false;
}
} catch (Exception e) {
return false;
}
}
private static String getPathJar() {
try {
final URI jarUriPath =
Application.class.getResource(Application.class.getSimpleName() + ".class").toURI();
String jarStringPath = jarUriPath.toString().replace("jar:", "");
String jarCleanPath = Paths.get(new URI(jarStringPath)).toString();
if (jarCleanPath.toLowerCase().contains(".jar")) {
return jarCleanPath.substring(0, jarCleanPath.lastIndexOf(".jar") + 4);
} else {
return null;
}
} catch (Exception e) {
return null;
}
}
}

Something that is frustrating is that when you are developing in Eclipse MyClass.class.getProtectionDomain().getCodeSource().getLocation() returns the /bin directory which is great, but when you compile it to a jar, the path includes the /myjarname.jar part which gives you illegal file names.
To have the code work both in the ide and once it is compiled to a jar, I use the following piece of code:
URL applicationRootPathURL = getClass().getProtectionDomain().getCodeSource().getLocation();
File applicationRootPath = new File(applicationRootPathURL.getPath());
File myFile;
if(applicationRootPath.isDirectory()){
myFile = new File(applicationRootPath, "filename");
}
else{
myFile = new File(applicationRootPath.getParentFile(), "filename");
}

Not really sure about the others but in my case it didn't work with a "Runnable jar" and i got it working by fixing codes together from phchen2 answer and another from this link :How to get the path of a running JAR file?
The code:
String path=new java.io.File(Server.class.getProtectionDomain()
.getCodeSource()
.getLocation()
.getPath())
.getAbsolutePath();
path=path.substring(0, path.lastIndexOf("."));
path=path+System.getProperty("java.class.path");

Have tried several of the solutions up there but none yielded correct results for the (probably special) case that the runnable jar has been exported with "Packaging external libraries" in Eclipse. For some reason all solutions based on the ProtectionDomain do result in null in that case.
From combining some solutions above I managed to achieve the following working code:
String surroundingJar = null;
// gets the path to the jar file if it exists; or the "bin" directory if calling from Eclipse
String jarDir = new File(ClassLoader.getSystemClassLoader().getResource(".").getPath()).getAbsolutePath();
// gets the "bin" directory if calling from eclipse or the name of the .jar file alone (without its path)
String jarFileFromSys = System.getProperty("java.class.path").split(";")[0];
// If both are equal that means it is running from an IDE like Eclipse
if (jarFileFromSys.equals(jarDir))
{
System.out.println("RUNNING FROM IDE!");
// The path to the jar is the "bin" directory in that case because there is no actual .jar file.
surroundingJar = jarDir;
}
else
{
// Combining the path and the name of the .jar file to achieve the final result
surroundingJar = jarDir + jarFileFromSys.substring(1);
}
System.out.println("JAR File: " + surroundingJar);

The above methods didn't work for me in my Spring environment, since Spring shades the actual classes into a package called BOOT-INF, thus not the actual location of the running file. I found another way to retrieve the running file through the Permissions object which have been granted to the running file:
public static Path getEnclosingDirectory() {
return Paths.get(FileUtils.class.getProtectionDomain().getPermissions()
.elements().nextElement().getName()).getParent();
}

Mention that it is checked only in Windows but i think it works perfect on other Operating Systems [Linux,MacOs,Solaris] :).
I had 2 .jar files in the same directory . I wanted from the one .jar file to start the other .jar file which is in the same directory.
The problem is that when you start it from the cmd the current directory is system32.
Warnings!
The below seems to work pretty well in all the test i have done even
with folder name ;][[;'57f2g34g87-8+9-09!2##!$%^^&() or ()%&$%^##
it works well.
I am using the ProcessBuilder with the below as following:
🍂..
//The class from which i called this was the class `Main`
String path = getBasePathForClass(Main.class);
String applicationPath= new File(path + "application.jar").getAbsolutePath();
System.out.println("Directory Path is : "+applicationPath);
//Your know try catch here
//Mention that sometimes it doesn't work for example with folder `;][[;'57f2g34g87-8+9-09!2##!$%^^&()`
ProcessBuilder builder = new ProcessBuilder("java", "-jar", applicationPath);
builder.redirectErrorStream(true);
Process process = builder.start();
//...code
🍂getBasePathForClass(Class<?> classs):
/**
* Returns the absolute path of the current directory in which the given
* class
* file is.
*
* #param classs
* #return The absolute path of the current directory in which the class
* file is.
* #author GOXR3PLUS[StackOverFlow user] + bachden [StackOverFlow user]
*/
public static final String getBasePathForClass(Class<?> classs) {
// Local variables
File file;
String basePath = "";
boolean failed = false;
// Let's give a first try
try {
file = new File(classs.getProtectionDomain().getCodeSource().getLocation().toURI().getPath());
if (file.isFile() || file.getPath().endsWith(".jar") || file.getPath().endsWith(".zip")) {
basePath = file.getParent();
} else {
basePath = file.getPath();
}
} catch (URISyntaxException ex) {
failed = true;
Logger.getLogger(classs.getName()).log(Level.WARNING,
"Cannot firgue out base path for class with way (1): ", ex);
}
// The above failed?
if (failed) {
try {
file = new File(classs.getClassLoader().getResource("").toURI().getPath());
basePath = file.getAbsolutePath();
// the below is for testing purposes...
// starts with File.separator?
// String l = local.replaceFirst("[" + File.separator +
// "/\\\\]", "")
} catch (URISyntaxException ex) {
Logger.getLogger(classs.getName()).log(Level.WARNING,
"Cannot firgue out base path for class with way (2): ", ex);
}
}
// fix to run inside eclipse
if (basePath.endsWith(File.separator + "lib") || basePath.endsWith(File.separator + "bin")
|| basePath.endsWith("bin" + File.separator) || basePath.endsWith("lib" + File.separator)) {
basePath = basePath.substring(0, basePath.length() - 4);
}
// fix to run inside netbeans
if (basePath.endsWith(File.separator + "build" + File.separator + "classes")) {
basePath = basePath.substring(0, basePath.length() - 14);
}
// end fix
if (!basePath.endsWith(File.separator)) {
basePath = basePath + File.separator;
}
return basePath;
}

This code worked for me:
private static String getJarPath() throws IOException, URISyntaxException {
File f = new File(LicensingApp.class.getProtectionDomain().().getLocation().toURI());
String jarPath = f.getCanonicalPath().toString();
String jarDir = jarPath.substring( 0, jarPath.lastIndexOf( File.separator ));
return jarDir;
}

The getProtectionDomain approach might not work sometimes e.g. when you have to find the jar for some of the core java classes (e.g in my case StringBuilder class within IBM JDK), however following works seamlessly:
public static void main(String[] args) {
System.out.println(findSource(MyClass.class));
// OR
System.out.println(findSource(String.class));
}
public static String findSource(Class<?> clazz) {
String resourceToSearch = '/' + clazz.getName().replace(".", "/") + ".class";
java.net.URL location = clazz.getResource(resourceToSearch);
String sourcePath = location.getPath();
// Optional, Remove junk
return sourcePath.replace("file:", "").replace("!" + resourceToSearch, "");
}

I have another way to get the String location of a class.
URL path = Thread.currentThread().getContextClassLoader().getResource("");
Path p = Paths.get(path.toURI());
String location = p.toString();
The output String will have the form of
C:\Users\Administrator\new Workspace\...
The spaces and other characters are handled, and in the form without file:/. So will be easier to use.

Rename All Files In Folders Java

I have one folder which contains many sub-folders. To make it more clear here is an example of the folders:
Movies:
MovieTitle:
Moviefile.mp4 (Movie File)
MovieSubtitles.srt (Subtitles)
MovieSeries:
MovieTitle:
Moviefile.mp4
MovieSubtitles.srt
I need to rename each mp4 and srt file to the following "MovieTitle". If the movie is part of a series it should be named to Series "Title + Movie Title". Lets use Star Wars as an example for series and how to name. "Star Wars" would be the name of a directory in "Movies". In "Star Wars" are 6 Folders each with a mp4 and srt file in it. For episode 1 of star wars the mp4 and srt file should be renamed to: "Star Wars - Episode 1.mp4" and "Star Wars - Episode 1.srt". If Episode 1 was not part of series it should be named to just "Episode 1.mp4"
Here is the code that I have come up with so far:
public static void renaming(File[] files){
String dir1, dir2;
for(File movie: files){ //Main folder containing all of the movies.
dir1 = movie.getName();
for(File filesInMovie: movie.listFiles()){
if(filesInMovie.isDirectory()){ //This means that it is a series.
dir2 = filesInMovie.getName();
for(File i: filesInMovie.listFiles()){
i.renameTo(dir1 + " - " + dir2);
}
}else{
filesInMovie.renameTo(dir1)
}
}
}
}
I realize that renameTo is an actual function in Java. I thought it would rename files until I read what it actually does (which I am still a little fuzzy on). So my main question is how would I get this code to properly rename the files.
Just some extra things you should know:
One Directory Contains all of the movies.
There are possibilities for each folder in the movies folder
It has other folders in it (It is a series)
It has a mp4 and srt file in it
If you have any questions please ask!!!

File#getName() returns a String. You are using this to get the name of the current file and then attempting to rename the file using File#rename(String) where the method is actually defined as File#renameTo(File). That is it the argument is expected to be a file to where you are attempting to rename the file.
Some Suggestions:
Look at Files.move() use this to move the file to target - Link has example of this
Don't use the get name method for the first directory - use the file directly (Something like below but still not correct I think) and this is if you want the file in the same base directory movies
i.renameTo(new File(movie," - " + filesInMovie.getName() + ext);
Still needs work for extension of file ext however which will come from the file i

Using Guava and Java 7, here is what you can do:
public final class Renamer
{
private static final Joiner JOINER = Joiner.on(" - ");
private static final class RenamerVisitor
extends SimpleFileVisitor<Path>
{
private final Path baseDir;
private RenamerVisitor(final Path baseDir)
{
this.baseDir = baseDir;
}
#Override
public FileVisitResult visitFile(final Path file,
final BasicFileAttributes attrs)
throws IOException
{
final Path relpath = baseDir.relativize(file);
final String targetName
= JOINER.join(Iterables.transform(relpath,
Functions.toStringFunction()));
final Path dstPath = baseDir.resolve(targetName);
System.out.printf("'%s' -> '%s'\n", file, dstPath);
return FileVisitResult.CONTINUE;
}
}
public static void main(final String... args)
throws IOException
{
final Path baseDir = Paths.get("/home/fge/tmp/jsr203/docs");
Files.walkFileTree(baseDir, new RenamerVisitor(baseDir));
}
}
This is for my environment in a directory. Sample output:
[...]
'/home/fge/tmp/jsr203/docs/java/io/class-use/File.html' -> '/home/fge/tmp/jsr203/docs/java - io - class-use - File.html'
'/home/fge/tmp/jsr203/docs/java/io/class-use/FilePermission.html' -> '/home/fge/tmp/jsr203/docs/java - io - class-use - FilePermission.html'
'/home/fge/tmp/jsr203/docs/java/io/File.html' -> '/home/fge/tmp/jsr203/docs/java - io - File.html'
In order to make it use the "real thing", replace the System.out.println() in visitFile with Files.move() and you're done!
Of course, you can also choose to copy instead, or even create a (hard) link if your filesystem supports it.
Side note: unlike File's .renameTo(), Files.move() does not silently fail!
(and after that, you should drop Java 6's old file API fast)