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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.
I want to create shortcuts inwindows with code, I use the library here:
http://alumnus.caltech.edu/~jimmc/jshortcut/jshortcut/README.html
also the corresponding codes:
import net.jimmc.jshortcut.JShellLink;
public class remove {
public static void main(String[] args) throws Exception{
String path = new String ("/home/test.csv");
readAndDelete(path, true, StandardCharsets.UTF_8);
}
private static void readAndDelete(String path, boolean ignoreHeader,Charset encoding) throws IOException {
File file = new File(path);
CSVParser parser = CSVParser.parse(file, encoding,CSVFormat.DEFAULT.withHeader());
List<CSVRecord> records = parser.getRecords();
List<String> docRecord = new ArrayList<String>();
List<String> shortpath = new ArrayList<String>();
for (CSVRecord doctype : records){
docRecord.add(doctype.get(0).toString());
shortpath.add(doctype.get(1).toString());
}
int recordlength = docRecord.size();
for(String eachdocRecord:docRecord){
try {
Path pathtemp=Paths.get(eachdocRecord);
Files.delete(pathtemp);
} catch (NoSuchFileException x) {
System.err.format("%s: no such" + " file or directory%n", path);
} catch (DirectoryNotEmptyException x) {
System.err.format("%s not empty%n", path);
} catch (IOException x) {
// File permission problems are caught here.
System.err.println(x);
}
}
for(int i=0; i<recordlength; i++){
JShellLink link = new JShellLink();
String pointpath=shortpath.get(i);
String originalpath = docRecord.get(i);
String[] parts = pointpath.split("\\\\");
int partssize= parts.length;
String name=parts[partssize-1];
String[] originalparts = originalpath.split("\\\\");
int originalsize = originalparts.length;
int lastlength = originalparts[originalsize-1].length();
String foldername = originalpath.substring(0,originalpath.length()-lastlength);
link.setFolder(foldername);
link.setName(name);
link.setPath(pointpath);
link.save();
}
}
}
I run it in windows command prompt, but always exception:
Exception in thread "main" java.lang.NoClassDefFoundError:net/jimmc/jshortcut/JShellLink
I compiled the .class successfully ...
Anyone could save nme ... thanks a lot
Regardless of anything else that might be wrong (I did not read the entire code) the exception is quite clear - JShellLink is not on your classpath. How to do it best depends on your use case - README suggests editing Manifest file when you build to .jar, it should also be possible to use Maven to take care of that, and any IDE should have been able to take care of classpaths for you. As far as I can tell however your use case looks something like
javac remove.java
java remove
(By the way, class names and class file names should start with upper case, that's the standard)
In that case the simplest way to do that would be to use:
java -cp .;jshortcut.jar remove
We add current directory (due to JShortcut wanting to have it's dll on classpath, just to be sure) as well as the jar containing classes you use to the classpath. If you're on Unix system use : instead of ;.
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.
i have a class which reads the list available in particular location,
the following is my code,
import java.io.File;
import java.io.IOException;
import java.util.ArrayList;
import java.util.List;
public class ExceptionInFileHandling {
#SuppressWarnings({ "rawtypes", "unchecked" })
public static void GetDirectory(String a_Path, List a_files, List a_folders) throws IOException {
try {
File l_Directory = new File(a_Path);
File[] l_files = l_Directory.listFiles();
for (int c = 0; c < l_files.length; c++) {
if (l_files[c].isDirectory()) {
a_folders.add(l_files[c].getName());
} else {
a_files.add(l_files[c].getName());
}
}
} catch (Exception ex){
ex.printStackTrace();
}
}
#SuppressWarnings("rawtypes")
public static void main(String args[]) throws IOException {
String filesLocation = "asdfasdf/sdfsdf/";
List l_Files = new ArrayList(), l_Folders = new ArrayList();
GetDirectory(filesLocation, l_Files, l_Folders);
System.out.println("Files");
System.out.println("---------------------------");
for (Object file : l_Files) {
System.out.println(file);
}
System.out.println("Done");
}
}
in this the file path can be passed as argument and that should be taken up based on the OS,
filePath.replaceAll("\\\\|/", "\\" + System.getProperty("file.separator"))
is this correct?
There are better ways to use file paths...
// Don't do this
filePath.replaceAll("\\\\|/", "\\" + System.getProperty("file.separator"))
Use java.nio.file.path:
import java.nio.file.*;
Path path = Paths.get(somePathString);
// Here is your system independent path
path.toAbsolutePath();
// Or this works too
Paths.get(somePathString).toAbsolutePath();
Use File.seperator:
// You can also input a String that has a proper file seperator like so
String filePath = "SomeDirectory" + File.separator;
// Then call your directory method
try{
ExceptionInFileHandling.GetDirectory(filePath, ..., ...);
} catch (Exception e){}
So a simple change to your method will now work cross platform:
#SuppressWarnings({ "rawtypes", "unchecked" })
public static void GetDirectory(String a_Path, List a_files, List a_folders) throws IOException {
try {
// File object is instead constructed
// with a URI by using Path.toUri()
// Change is done here
File l_Directory = new File(Paths.get(a_Path).toUri());
File[] l_files = l_Directory.listFiles();
for (int c = 0; c < l_files.length; c++) {
if (l_files[c].isDirectory()) {
a_folders.add(l_files[c].getName());
} else {
a_files.add(l_files[c].getName());
}
}
} catch (Exception ex){
ex.printStackTrace();
}
}
You can use forward slashes as directory separators on Windows as well when calling File constructor.
why you are not adding java defined file separator instead of creating a string then replacing all.
try it like
String filesLocation = "asdfasdf"+File.separator+"sdfsdf"+File.separator;
The answer from you should be correct. There is another similiar thread with answer :
Java regex to replace file path based on OS
Platform independent paths in Java
You could generate a Path object from the passed argument. Then you would not need to handle the file separator on your own.
public static void main(String[] args) {
Path path = Paths.get(args[0]);
System.out.println("path = " + path.toAbsolutePath());
}
The code is able to handle following passed arguments.
foo\bar
foo/bar
foo\bar/baz
foo\\bar
foo//baz
foo\\bar//baz
...
First, you should not used relative path like asdfasdf/sdfsdf/. It's a big source of bugs as your path depends on your working directory.
That thing said, your replaceAll is quite good but it can be improved like this :
filePath.replaceAll(
"[/\\\\]+",
Matcher.quoteReplacement(System.getProperty("file.separator")));
Using quoteReplacement is adviced in replaceAll documentation
Returns a literal replacement String for the specified String. This method produces a String that will work as a literal replacement s in the appendReplacement method of the Matcher class. The String produced will match the sequence of characters in s treated as a literal sequence. Slashes ('\') and dollar signs ('$') will be given no special meaning.
You need to use java.io.File.separatorChar to The system-dependent default name-separator character.
String location = "usr"+java.io.File.separatorChar+"local"+java.io.File.separatorChar;
org.apache.commons.io.FilenameUtils contains a lot of useful methods, for instance separatorsToSystem(String path) converts separators in given path in accordance with OS which you are using.
Use the below method to know about OS file separator and then replaceAll previous separator with this methods.
System.getProperty("file.separator");
Why you just dont use "/". It is acceptable for both linux and windows as path seperator.
I would like to check whether an existing file is in a specific directory or a subdirectory of that.
I have two File objects.
File dir;
File file;
Both are guaranteed to exist. Let's assume
dir = /tmp/dir
file = /tmp/dir/subdir1/subdir2/file.txt
I want this check to return true
For now i am doing the check this way:
String canonicalDir = dir.getCanonicalPath() + File.separator;
boolean subdir = file.getCanonicalPath().startsWith(canonicalDir);
This seems to work with my limited tests, but i am unsure whether this might make problems on some operating systems. I also do not like that getCanonicalPath() can throw an IOException which i have to handle.
Is there a better way? Possibly in some library?
Thanks
In addition to the asnwer from rocketboy, use getCanonicalPath() instad of getAbsolutePath() so \dir\dir2\..\file is converted to \dir\file:
boolean areRelated = file.getCanonicalPath().contains(dir.getCanonicalPath() + File.separator);
System.out.println(areRelated);
or
boolean areRelated = child.getCanonicalPath().startsWith(parent.getCanonicalPath() + File.separator);
Do not forget to catch any Exception with try {...} catch {...}.
NOTE: You can use FileSystem.getSeparator() instead of File.separator. The 'correct' way of doing this will be to get the getCanonicalPath() of the directory that you are going to check against as a String, then check if ends with a File.separator and if not then add File.separator to the end of that String, to avoid double slashes. This way you skip future odd behaviours if Java decides to return directories with a slash in the end or if your directory string comes from somewhere else than Java.io.File.
NOTE2: Thanx to #david for pointing the File.separator problem.
I would create a small utility method:
public static boolean isInSubDirectory(File dir, File file) {
if (file == null)
return false;
if (file.equals(dir))
return true;
return isInSubDirectory(dir, file.getParentFile());
}
This method looks pretty solid:
/**
* Checks, whether the child directory is a subdirectory of the base
* directory.
*
* #param base the base directory.
* #param child the suspected child directory.
* #return true, if the child is a subdirectory of the base directory.
* #throws IOException if an IOError occured during the test.
*/
public boolean isSubDirectory(File base, File child)
throws IOException {
base = base.getCanonicalFile();
child = child.getCanonicalFile();
File parentFile = child;
while (parentFile != null) {
if (base.equals(parentFile)) {
return true;
}
parentFile = parentFile.getParentFile();
}
return false;
}
Source
It is similar to the solution by dacwe but doesn't use recursion (though that shouldn't make a big difference in this case).
If you plan to works with file and filenames heavly check apache fileutils and filenameutils libraries. Are full of useful (and portale if portability is mamdatory) functions
public class Test {
public static void main(String[] args) {
File root = new File("c:\\test");
String fileName = "a.txt";
try {
boolean recursive = true;
Collection files = FileUtils.listFiles(root, null, recursive);
for (Iterator iterator = files.iterator(); iterator.hasNext();) {
File file = (File) iterator.next();
if (file.getName().equals(fileName))
System.out.println(file.getAbsolutePath());
}
} catch (Exception e) {
e.printStackTrace();
}
}
}
You can traverse File Tree starting from your specific DIR.
At Java 7, there is Files.walkFileTree method. You have only to write your own visitor
to check if current node is searched file. More doc:
http://docs.oracle.com/javase/7/docs/api/java/nio/file/Files.html#walkFileTree%28java.nio.file.Path,%20java.util.Set,%20int,%20java.nio.file.FileVisitor%29
You can do this, however it won't catch every use case e.g. dir = /somedir/../tmp/dir/etc..., unless that's how the file was defined also.
import java.nio.file.Path;
import java.nio.file.Paths;
public class FileTest {
public static void main(final String... args) {
final Path dir = Paths.get("/tmp/dir").toAbsolutePath();
final Path file = Paths.get("/tmp/dir/subdir1/subdir2/file.txt").toAbsolutePath();
System.out.println("Dir: " + dir);
System.out.println("File: " + file);
final boolean valid = file.startsWith(dir);
System.out.println("Valid: " + valid);
}
}
In order for the checks to work correctly, you really need to map these using toRealPath() or, in your example, getCanonicalPath(), but you then have to handle exceptions for these examples which is absolutely correct that you should do so.
Since Java 7+ you can just do this:
file.toPath().startsWith(dir.toPath());
How about comparing the paths?
boolean areRelated = file.getAbsolutePath().contains(dir.getAbsolutePath());
System.out.println(areRelated);
or
boolean areRelated = child.getAbsolutePath().startsWith(parent.getAbsolutePath())