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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'm writing a Java program which will execute an external file ~/Java/exampleProject/bin/import.sh. My program is under package gqqnbig. So the directory structure is
exampleProject/
bin/
import.sh
gqqnbig/
*.class
When I debug the program in eclipse, the working directory is ~/Java/exampleProject/. I have to execute bin/import.sh.
When I run the program in cmd, the current directory is ~/Java/exampleProject/bin, my code will not find import.sh.
The program has to be portable (distribute with import.sh). With the correct directory structure, it should work in my computer as well as in your computer, so I cannot hard code the path of import.sh.
I also want to pack it into a single jar file. The desired structure is (Figure 1)
bin/
import.sh
program.jar
So how can my program find import.sh when run in eclipse, cmd and jar?
UPDATE
I ask my question in another way. Please implement getAbsolutePath function, so that no matther the code is running in eclipse, in cmd, or as a jar file in a folder which also has import.sh (See Figure 1), the output is identical.
public static void main(String[] args)
{
System.out.println("Look for "+getAbsolutePath()+"\\import.sh");
}
Here's a method pulled from one of my projects. It get's the folder that the jar file is located in as opposed to the directory if was run from if invoked on the command line.
/**
* Retrieve a File representation of the folder this application is
* located in.
*
* #return
*/
private static File getApplicationRootFolder()
{
String path = FileGetter.class.getProtectionDomain().getCodeSource()
.getLocation().getPath();
try
{
String decodedPath = URLDecoder.decode(path, "UTF-8");
File jarParentFolder = new File(decodedPath).getParentFile();
if (jarParentFolder.exists() && jarParentFolder.canRead()
{
File shellScript = new File(jarParentFolder, "import.sh")
}
catch (UnsupportedEncodingException e)
{
Main.myLog.error(TAG, "Unencoding jar path failed on:\n\t" + path);
e.printStackTrace();
return null;
}
}
You can then use that directory to make a File object for your shell script File shellScript = new File(getApplicationRootFolder(), scriptFilename);
EDIT: Follow up questions to try to help you out and a solution
So you want to be able to access one file that has three locations depending on when/where you code is run. This is how I see those cases:
Case 1: Running directly from Eclipse (unpackaged code):
shell script: X:/Java/exampleProject/bin/import.sh
class file: X:/Java/exampleProject/bin/gqqnbig/YourClass.class
Case 2: Running the packaged jar (shell script inside):
shell script: X:/Java/YourJar.jar/bin/import.sh
class file: X:/Java/YourJar.jar/bin/gqqnbig/YourClass.class
Case 3: Running the packaged jar (shell script external):
shell script: X:/Java/import.sh
class file: X:/Java/YourJar.jar/bin/gqqnbig/YourClass.class
What I think you need to do is prioritise the order you look at these locations and fall back to the next one in line if the shell script isn't found. I'd guess you want:
1. external to jar
2. inside packaged jar
3. unpackaged
So to access these you will need to write each separately and move through each until you get File.exists() == true.
Something like what follows. Note I didn't test this and there are likely errors. I'll leave you to sort them out. My code is based on the assumptions made above, again I'll leave you to modify the code based on any incorrect guesses.
So here's a class with one public method taking a filename argument and returning an InputStream. I opted for InputStream in all cases as once you package up your jar you cannot access the resources as File objects any more, only Streams.
public class FileGetter
{
private static String RESOURCE_DIRECTORY = "bin";
/**
* Retrieve an InputStream for a resource file.
*
* #param filename
* #return
*/
public InputStream getResourceFileStream(String filename)
{
// this is where you decide your preference or the priority of the locations
InputStream inputStream = null;
inputStream = getExternalFile(filename);
if (inputStream != null)
{
return inputStream;
}
inputStream = getInternalPackagedFile(filename);
if (inputStream != null)
{
return inputStream;
}
inputStream = getInternalUnpackagedFile(filename);
if (inputStream != null)
{
return inputStream;
}
// couldn't find the file anywhere so log some error or throw an exception
return null;
}
/**
* Retrieve an InputStream for a file located outside your Jar
*
* #param filename
* #return
*/
private static InputStream getExternalFile(String filename)
{
// get the jar's absolute location on disk (regardless of current 'working directory')
String appRootPath = FileGetter.class.getProtectionDomain().getCodeSource()
.getLocation().getPath();
try
{
String decodedPath = URLDecoder.decode(appRootPath, "UTF-8");
File jarfile = new File(decodedPath);
File parentDirectory = jarfile.getParentFile();
if (testExists(parentDirectory))
{
File shellScript = new File(parentDirectory, filename);
if (testExists(shellScript))
{
return new FileInputStream(shellScript);
}
}
}
catch (UnsupportedEncodingException e)
{}
catch (NullPointerException e)
{}
catch (FileNotFoundException e)
{}
// if any part fails return null
return null;
}
/**
* Retrieve an InputStream for a file located inside your Jar.
*
* #param filename
* #return
*/
private static InputStream getInternalPackagedFile(String filename)
{
// root directory is defined as the jar's root so we start with a "/".
URL resUrl = FileGetter.class.getResource(File.separator + RESOURCE_DIRECTORY
+ File.separator + filename);
String badPath = resUrl.getPath();
String goodPath = badPath.substring(badPath.indexOf("!") + 1);
InputStream input = FileGetter.class.getResourceAsStream(goodPath);
// returns null if nothing there so just
return input;
}
private static InputStream getInternalUnpackagedFile(String filename)
{
// eclipse will 'cd' to the code's directory so we use relative paths
File shellScriptFile = new File(RESOURCE_DIRECTORY + File.separator + filename);
if (testExists(shellScriptFile))
{
try
{
InputStream shellScriptStream = new FileInputStream(shellScriptFile);
if (shellScriptStream != null)
{
return shellScriptStream;
}
}
catch (FileNotFoundException e)
{}
}
return null;
}
/**
* Test that a file exists and can be read.
*
* #param file
* #return
*/
private static boolean testExists(File file)
{
return file != null && file.exists() && file.canRead();
}
}
But with all that being said a better way to sort this would be to ensure that the file exists on disk and create it if not found. Then execute the script from disk.
I would like to know a definitive answer for this myself.
As a workaround I would put 'import.sh' inside the exampleProject and change the relative path to 'import.sh'.
In theory that should work inside Eclipse, and as a packaged Jar file with program.jar and import.sh in the same directory.
It won't work on the cmd prompt unfortunately, maybe someone can suggest a better method.
-Kaz
I composed a solution. Call getExecutablePath() to get unified path.
public static File getExecutablePath()
{
String workingDirectory = System.getProperty("user.dir");
File binFile = new File(workingDirectory, "bin");
if (binFile.exists() && (new File(workingDirectory, "src")).exists())
{
return binFile;
}
else if (isRunningFromJar())
return getApplicationRootFolder();
else
return new File(workingDirectory);
}
public static boolean isRunningFromJar()
{
String className = SystemHelper.class.getName().replace('.', '/');
String classJar = SystemHelper.class.getResource("/" + className + ".class").toString();
return classJar.startsWith("jar:");
}
/**
* Retrieve a File representation of the folder this application is located in.
*
* #return
*/
private static File getApplicationRootFolder()
{
try
{
String path = SystemHelper.class.getProtectionDomain().getCodeSource().getLocation().getPath();
String decodedPath;
decodedPath = URLDecoder.decode(path, "UTF-8");
File jarfile = new File(decodedPath);
return jarfile.getParentFile();
}
catch (UnsupportedEncodingException e)
{
throw new RuntimeException(e);
}
}
I want to copy a file from a jar. The file that I am copying is going to be copied outside the working directory. I have done some tests and all methods I try end up with 0 byte files.
EDIT: I want the copying of the file to be done via a program, not manually.
First of all I want to say that some answers posted before are entirely correct, but I want to give mine, since sometimes we can't use open source libraries under the GPL, or because we are too lazy to download the jar XD or what ever your reason is here is a standalone solution.
The function below copy the resource beside the Jar file:
/**
* Export a resource embedded into a Jar file to the local file path.
*
* #param resourceName ie.: "/SmartLibrary.dll"
* #return The path to the exported resource
* #throws Exception
*/
static public String ExportResource(String resourceName) throws Exception {
InputStream stream = null;
OutputStream resStreamOut = null;
String jarFolder;
try {
stream = ExecutingClass.class.getResourceAsStream(resourceName);//note that each / is a directory down in the "jar tree" been the jar the root of the tree
if(stream == null) {
throw new Exception("Cannot get resource \"" + resourceName + "\" from Jar file.");
}
int readBytes;
byte[] buffer = new byte[4096];
jarFolder = new File(ExecutingClass.class.getProtectionDomain().getCodeSource().getLocation().toURI().getPath()).getParentFile().getPath().replace('\\', '/');
resStreamOut = new FileOutputStream(jarFolder + resourceName);
while ((readBytes = stream.read(buffer)) > 0) {
resStreamOut.write(buffer, 0, readBytes);
}
} catch (Exception ex) {
throw ex;
} finally {
stream.close();
resStreamOut.close();
}
return jarFolder + resourceName;
}
Just change ExecutingClass to the name of your class, and call it like this:
String fullPath = ExportResource("/myresource.ext");
Edit for Java 7+ (for your convenience)
As answered by GOXR3PLUS and noted by Andy Thomas you can achieve this with:
Files.copy( InputStream in, Path target, CopyOption... options)
See GOXR3PLUS answer for more details
Given your comment about 0-byte files, I have to assume you're trying to do this programmatically, and, given your tags, that you're doing it in Java. If that's true, then just use Class.getResource() to get a URL pointing to the file in your JAR, then Apache Commons IO FileUtils.copyURLToFile() to copy it out to the file system. E.g.:
URL inputUrl = getClass().getResource("/absolute/path/of/source/in/jar/file");
File dest = new File("/path/to/destination/file");
FileUtils.copyURLToFile(inputUrl, dest);
Most likely, the problem with whatever code you have now is that you're (correctly) using a buffered output stream to write to the file but (incorrectly) failing to close it.
Oh, and you should edit your question to clarify exactly how you want to do this (programmatically, not, language, ...)
Faster way to do it with Java 7+ , plus code to get the current directory:
/**
* Copy a file from source to destination.
*
* #param source
* the source
* #param destination
* the destination
* #return True if succeeded , False if not
*/
public static boolean copy(InputStream source , String destination) {
boolean succeess = true;
System.out.println("Copying ->" + source + "\n\tto ->" + destination);
try {
Files.copy(source, Paths.get(destination), StandardCopyOption.REPLACE_EXISTING);
} catch (IOException ex) {
logger.log(Level.WARNING, "", ex);
succeess = false;
}
return succeess;
}
Testing it (icon.png is an image inside the package image of the application):
copy(getClass().getResourceAsStream("/image/icon.png"),getBasePathForClass(Main.class)+"icon.png");
About the line of code (getBasePathForClass(Main.class)): -> check the answer i have added here :) -> Getting the Current Working Directory in Java
Java 8 (actually FileSystem is there since 1.7) comes with some cool new classes/methods to deal with this. As somebody already mentioned that JAR is basically ZIP file, you could use
final URI jarFileUril = URI.create("jar:file:" + file.toURI().getPath());
final FileSystem fs = FileSystems.newFileSystem(jarFileUri, env);
(See Zip File)
Then you can use one of the convenient methods like:
fs.getPath("filename");
Then you can use Files class
try (final Stream<Path> sources = Files.walk(from)) {
sources.forEach(src -> {
final Path dest = to.resolve(from.relativize(src).toString());
try {
if (Files.isDirectory(from)) {
if (Files.notExists(to)) {
log.trace("Creating directory {}", to);
Files.createDirectories(to);
}
} else {
log.trace("Extracting file {} to {}", from, to);
Files.copy(from, to, StandardCopyOption.REPLACE_EXISTING);
}
} catch (IOException e) {
throw new RuntimeException("Failed to unzip file.", e);
}
});
}
Note: I tried that to unpack JAR files for testing
Robust solution:
public static void copyResource(String res, String dest, Class c) throws IOException {
InputStream src = c.getResourceAsStream(res);
Files.copy(src, Paths.get(dest), StandardCopyOption.REPLACE_EXISTING);
}
You can use it like this:
File tempFileGdalZip = File.createTempFile("temp_gdal", ".zip");
copyResource("/gdal.zip", tempFileGdalZip.getAbsolutePath(), this.getClass());
Use the JarInputStream class:
// assuming you already have an InputStream to the jar file..
JarInputStream jis = new JarInputStream( is );
// get the first entry
JarEntry entry = jis.getNextEntry();
// we will loop through all the entries in the jar file
while ( entry != null ) {
// test the entry.getName() against whatever you are looking for, etc
if ( matches ) {
// read from the JarInputStream until the read method returns -1
// ...
// do what ever you want with the read output
// ...
// if you only care about one file, break here
}
// get the next entry
entry = jis.getNextEntry();
}
jis.close();
See also: JarEntry
To copy a file from your jar, to the outside, you need to use the following approach:
Get a InputStream to a the file inside your jar file using getResourceAsStream()
We open our target file using a FileOutputStream
We copy bytes from the input to the output stream
We close our streams to prevent resource leaks
Example code that also contains a variable to not replace the existing values:
public File saveResource(String name) throws IOException {
return saveResource(name, true);
}
public File saveResource(String name, boolean replace) throws IOException {
return saveResource(new File("."), name, replace)
}
public File saveResource(File outputDirectory, String name) throws IOException {
return saveResource(outputDirectory, name, true);
}
public File saveResource(File outputDirectory, String name, boolean replace)
throws IOException {
File out = new File(outputDirectory, name);
if (!replace && out.exists())
return out;
// Step 1:
InputStream resource = this.getClass().getResourceAsStream(name);
if (resource == null)
throw new FileNotFoundException(name + " (resource not found)");
// Step 2 and automatic step 4
try(InputStream in = resource;
OutputStream writer = new BufferedOutputStream(
new FileOutputStream(out))) {
// Step 3
byte[] buffer = new byte[1024 * 4];
int length;
while((length = in.read(buffer)) >= 0) {
writer.write(buffer, 0, length);
}
}
return out;
}
A jar is just a zip file. Unzip it (using whatever method you're comfortable with) and copy the file normally.
${JAVA_HOME}/bin/jar -cvf /path/to.jar
Is there a standard and reliable way of creating a temporary directory inside a Java application? There's an entry in Java's issue database, which has a bit of code in the comments, but I wonder if there is a standard solution to be found in one of the usual libraries (Apache Commons etc.) ?
If you are using JDK 7 use the new Files.createTempDirectory class to create the temporary directory.
Path tempDirWithPrefix = Files.createTempDirectory(prefix);
Before JDK 7 this should do it:
public static File createTempDirectory()
throws IOException
{
final File temp;
temp = File.createTempFile("temp", Long.toString(System.nanoTime()));
if(!(temp.delete()))
{
throw new IOException("Could not delete temp file: " + temp.getAbsolutePath());
}
if(!(temp.mkdir()))
{
throw new IOException("Could not create temp directory: " + temp.getAbsolutePath());
}
return (temp);
}
You could make better exceptions (subclass IOException) if you want.
The Google Guava library has a ton of helpful utilities. One of note here is the Files class. It has a bunch of useful methods including:
File myTempDir = Files.createTempDir();
This does exactly what you asked for in one line. If you read the documentation here you'll see that the proposed adaptation of File.createTempFile("install", "dir") typically introduces security vulnerabilities.
If you need a temporary directory for testing and you are using jUnit, #Rule together with TemporaryFolder solves your problem:
#Rule
public TemporaryFolder folder = new TemporaryFolder();
From the documentation:
The TemporaryFolder Rule allows creation of files and folders that are guaranteed to be deleted when the test method finishes (whether it passes or fails)
Update:
If you are using JUnit Jupiter (version 5.1.1 or greater), you have the option to use JUnit Pioneer which is the JUnit 5 Extension Pack.
Copied from the project documentation:
For example, the following test registers the extension for a single test method, creates and writes a file to the temporary directory and checks its content.
#Test
#ExtendWith(TempDirectory.class)
void test(#TempDir Path tempDir) {
Path file = tempDir.resolve("test.txt");
writeFile(file);
assertExpectedFileContent(file);
}
More info in the JavaDoc and the JavaDoc of TempDirectory
Gradle:
dependencies {
testImplementation 'org.junit-pioneer:junit-pioneer:0.1.2'
}
Maven:
<dependency>
<groupId>org.junit-pioneer</groupId>
<artifactId>junit-pioneer</artifactId>
<version>0.1.2</version>
<scope>test</scope>
</dependency>
Update 2:
The #TempDir annotation was added to the JUnit Jupiter 5.4.0 release as an experimental feature. Example copied from the JUnit 5 User Guide:
#Test
void writeItemsToFile(#TempDir Path tempDir) throws IOException {
Path file = tempDir.resolve("test.txt");
new ListWriter(file).write("a", "b", "c");
assertEquals(singletonList("a,b,c"), Files.readAllLines(file));
}
Naively written code to solve this problem suffers from race conditions, including several of the answers here. Historically you could think carefully about race conditions and write it yourself, or you could use a third-party library like Google's Guava (as Spina's answer suggested.) Or you could write buggy code.
But as of JDK 7, there is good news! The Java standard library itself now provides a properly working (non-racy) solution to this problem. You want java.nio.file.Files#createTempDirectory(). From the documentation:
public static Path createTempDirectory(Path dir,
String prefix,
FileAttribute<?>... attrs)
throws IOException
Creates a new directory in the specified directory, using the given prefix to generate its name. The resulting Path is associated with the same FileSystem as the given directory.
The details as to how the name of the directory is constructed is implementation dependent and therefore not specified. Where possible the prefix is used to construct candidate names.
This effectively resolves the embarrassingly ancient bug report in the Sun bug tracker which asked for just such a function.
This is the source code to the Guava library's Files.createTempDir(). It's nowhere as complex as you might think:
public static File createTempDir() {
File baseDir = new File(System.getProperty("java.io.tmpdir"));
String baseName = System.currentTimeMillis() + "-";
for (int counter = 0; counter < TEMP_DIR_ATTEMPTS; counter++) {
File tempDir = new File(baseDir, baseName + counter);
if (tempDir.mkdir()) {
return tempDir;
}
}
throw new IllegalStateException("Failed to create directory within "
+ TEMP_DIR_ATTEMPTS + " attempts (tried "
+ baseName + "0 to " + baseName + (TEMP_DIR_ATTEMPTS - 1) + ')');
}
By default:
private static final int TEMP_DIR_ATTEMPTS = 10000;
See here
Do not use deleteOnExit() even if you explicitly delete it later.
Google 'deleteonexit is evil' for more info, but the gist of the problem is:
deleteOnExit() only deletes for normal JVM shutdowns, not crashes or killing the JVM process.
deleteOnExit() only deletes on JVM shutdown - not good for long running server processes because:
The most evil of all - deleteOnExit() consumes memory for each temp file entry. If your process is running for months, or creates a lot of temp files in a short time, you consume memory and never release it until the JVM shuts down.
As of Java 1.7 createTempDirectory(prefix, attrs) and createTempDirectory(dir, prefix, attrs) are included in java.nio.file.Files
Example:
File tempDir = Files.createTempDirectory("foobar").toFile();
This is what I decided to do for my own code:
/**
* Create a new temporary directory. Use something like
* {#link #recursiveDelete(File)} to clean this directory up since it isn't
* deleted automatically
* #return the new directory
* #throws IOException if there is an error creating the temporary directory
*/
public static File createTempDir() throws IOException
{
final File sysTempDir = new File(System.getProperty("java.io.tmpdir"));
File newTempDir;
final int maxAttempts = 9;
int attemptCount = 0;
do
{
attemptCount++;
if(attemptCount > maxAttempts)
{
throw new IOException(
"The highly improbable has occurred! Failed to " +
"create a unique temporary directory after " +
maxAttempts + " attempts.");
}
String dirName = UUID.randomUUID().toString();
newTempDir = new File(sysTempDir, dirName);
} while(newTempDir.exists());
if(newTempDir.mkdirs())
{
return newTempDir;
}
else
{
throw new IOException(
"Failed to create temp dir named " +
newTempDir.getAbsolutePath());
}
}
/**
* Recursively delete file or directory
* #param fileOrDir
* the file or dir to delete
* #return
* true iff all files are successfully deleted
*/
public static boolean recursiveDelete(File fileOrDir)
{
if(fileOrDir.isDirectory())
{
// recursively delete contents
for(File innerFile: fileOrDir.listFiles())
{
if(!FileUtilities.recursiveDelete(innerFile))
{
return false;
}
}
}
return fileOrDir.delete();
}
Well, "createTempFile" actually creates the file. So why not just delete it first, and then do the mkdir on it?
This code should work reasonably well:
public static File createTempDir() {
final String baseTempPath = System.getProperty("java.io.tmpdir");
Random rand = new Random();
int randomInt = 1 + rand.nextInt();
File tempDir = new File(baseTempPath + File.separator + "tempDir" + randomInt);
if (tempDir.exists() == false) {
tempDir.mkdir();
}
tempDir.deleteOnExit();
return tempDir;
}
As discussed in this RFE and its comments, you could call tempDir.delete() first. Or you could use System.getProperty("java.io.tmpdir") and create a directory there. Either way, you should remember to call tempDir.deleteOnExit(), or the file won't be deleted after you're done.
Just for completion, this is the code from google guava library. It is not my code, but I think it is valueable to show it here in this thread.
/** Maximum loop count when creating temp directories. */
private static final int TEMP_DIR_ATTEMPTS = 10000;
/**
* Atomically creates a new directory somewhere beneath the system's temporary directory (as
* defined by the {#code java.io.tmpdir} system property), and returns its name.
*
* <p>Use this method instead of {#link File#createTempFile(String, String)} when you wish to
* create a directory, not a regular file. A common pitfall is to call {#code createTempFile},
* delete the file and create a directory in its place, but this leads a race condition which can
* be exploited to create security vulnerabilities, especially when executable files are to be
* written into the directory.
*
* <p>This method assumes that the temporary volume is writable, has free inodes and free blocks,
* and that it will not be called thousands of times per second.
*
* #return the newly-created directory
* #throws IllegalStateException if the directory could not be created
*/
public static File createTempDir() {
File baseDir = new File(System.getProperty("java.io.tmpdir"));
String baseName = System.currentTimeMillis() + "-";
for (int counter = 0; counter < TEMP_DIR_ATTEMPTS; counter++) {
File tempDir = new File(baseDir, baseName + counter);
if (tempDir.mkdir()) {
return tempDir;
}
}
throw new IllegalStateException(
"Failed to create directory within "
+ TEMP_DIR_ATTEMPTS
+ " attempts (tried "
+ baseName
+ "0 to "
+ baseName
+ (TEMP_DIR_ATTEMPTS - 1)
+ ')');
}
I got the same problem so this is just another answer for those who are interested, and it's similar to one of the above:
public static final String tempDir = System.getProperty("java.io.tmpdir")+"tmp"+System.nanoTime();
static {
File f = new File(tempDir);
if(!f.exists())
f.mkdir();
}
And for my application, I decided that to add in a option to clear the temp on exit so I added in a shut-down hook:
Runtime.getRuntime().addShutdownHook(new Thread() {
#Override
public void run() {
//stackless deletion
String root = MainWindow.tempDir;
Stack<String> dirStack = new Stack<String>();
dirStack.push(root);
while(!dirStack.empty()) {
String dir = dirStack.pop();
File f = new File(dir);
if(f.listFiles().length==0)
f.delete();
else {
dirStack.push(dir);
for(File ff: f.listFiles()) {
if(ff.isFile())
ff.delete();
else if(ff.isDirectory())
dirStack.push(ff.getPath());
}
}
}
}
});
The method delete all subdirs and files before deleting the temp, without using the callstack (which is totally optional and you could do it with recursion at this point), but I want to be on the safe side.
As you can see in the other answers, no standard approach has arisen.
Hence you already mentioned Apache Commons, I propose the following approach using FileUtils from Apache Commons IO:
/**
* Creates a temporary subdirectory in the standard temporary directory.
* This will be automatically deleted upon exit.
*
* #param prefix
* the prefix used to create the directory, completed by a
* current timestamp. Use for instance your application's name
* #return the directory
*/
public static File createTempDirectory(String prefix) {
final File tmp = new File(FileUtils.getTempDirectory().getAbsolutePath()
+ "/" + prefix + System.currentTimeMillis());
tmp.mkdir();
Runtime.getRuntime().addShutdownHook(new Thread() {
#Override
public void run() {
try {
FileUtils.deleteDirectory(tmp);
} catch (IOException e) {
e.printStackTrace();
}
}
});
return tmp;
}
This is preferred since apache commons the library that comes as closest to the asked "standard" and works with both JDK 7 and older versions. This also returns an "old" File instance (which is stream based) and not a "new" Path instance (which is buffer based and would be the result of JDK7's getTemporaryDirectory() method) -> Therefore it returns what most people need when they want to create a temporary directory.
Try this small example:
Code:
try {
Path tmpDir = Files.createTempDirectory("tmpDir");
System.out.println(tmpDir.toString());
Files.delete(tmpDir);
} catch (IOException e) {
e.printStackTrace();
}
Imports:
java.io.IOException
java.nio.file.Files
java.nio.file.Path
Console output on Windows machine:
C:\Users\userName\AppData\Local\Temp\tmpDir2908538301081367877
Comment:
Files.createTempDirectory generates unique ID atomatically - 2908538301081367877.
Note:
Read the following for deleting directories recursively:
Delete directories recursively in Java
I like the multiple attempts at creating a unique name but even this solution does not rule out a race condition. Another process can slip in after the test for exists() and the if(newTempDir.mkdirs()) method invocation. I have no idea how to completely make this safe without resorting to native code, which I presume is what's buried inside File.createTempFile().
Before Java 7 you could also:
File folder = File.createTempFile("testFileUtils", ""); // no suffix
folder.delete();
folder.mkdirs();
folder.deleteOnExit();
Using File#createTempFile and delete to create a unique name for the directory seems ok. You should add a ShutdownHook to delete the directory (recursively) on JVM shutdown.
Is there a standard and reliable way of creating a temporary directory inside a Java application? There's an entry in Java's issue database, which has a bit of code in the comments, but I wonder if there is a standard solution to be found in one of the usual libraries (Apache Commons etc.) ?
If you are using JDK 7 use the new Files.createTempDirectory class to create the temporary directory.
Path tempDirWithPrefix = Files.createTempDirectory(prefix);
Before JDK 7 this should do it:
public static File createTempDirectory()
throws IOException
{
final File temp;
temp = File.createTempFile("temp", Long.toString(System.nanoTime()));
if(!(temp.delete()))
{
throw new IOException("Could not delete temp file: " + temp.getAbsolutePath());
}
if(!(temp.mkdir()))
{
throw new IOException("Could not create temp directory: " + temp.getAbsolutePath());
}
return (temp);
}
You could make better exceptions (subclass IOException) if you want.
The Google Guava library has a ton of helpful utilities. One of note here is the Files class. It has a bunch of useful methods including:
File myTempDir = Files.createTempDir();
This does exactly what you asked for in one line. If you read the documentation here you'll see that the proposed adaptation of File.createTempFile("install", "dir") typically introduces security vulnerabilities.
If you need a temporary directory for testing and you are using jUnit, #Rule together with TemporaryFolder solves your problem:
#Rule
public TemporaryFolder folder = new TemporaryFolder();
From the documentation:
The TemporaryFolder Rule allows creation of files and folders that are guaranteed to be deleted when the test method finishes (whether it passes or fails)
Update:
If you are using JUnit Jupiter (version 5.1.1 or greater), you have the option to use JUnit Pioneer which is the JUnit 5 Extension Pack.
Copied from the project documentation:
For example, the following test registers the extension for a single test method, creates and writes a file to the temporary directory and checks its content.
#Test
#ExtendWith(TempDirectory.class)
void test(#TempDir Path tempDir) {
Path file = tempDir.resolve("test.txt");
writeFile(file);
assertExpectedFileContent(file);
}
More info in the JavaDoc and the JavaDoc of TempDirectory
Gradle:
dependencies {
testImplementation 'org.junit-pioneer:junit-pioneer:0.1.2'
}
Maven:
<dependency>
<groupId>org.junit-pioneer</groupId>
<artifactId>junit-pioneer</artifactId>
<version>0.1.2</version>
<scope>test</scope>
</dependency>
Update 2:
The #TempDir annotation was added to the JUnit Jupiter 5.4.0 release as an experimental feature. Example copied from the JUnit 5 User Guide:
#Test
void writeItemsToFile(#TempDir Path tempDir) throws IOException {
Path file = tempDir.resolve("test.txt");
new ListWriter(file).write("a", "b", "c");
assertEquals(singletonList("a,b,c"), Files.readAllLines(file));
}
Naively written code to solve this problem suffers from race conditions, including several of the answers here. Historically you could think carefully about race conditions and write it yourself, or you could use a third-party library like Google's Guava (as Spina's answer suggested.) Or you could write buggy code.
But as of JDK 7, there is good news! The Java standard library itself now provides a properly working (non-racy) solution to this problem. You want java.nio.file.Files#createTempDirectory(). From the documentation:
public static Path createTempDirectory(Path dir,
String prefix,
FileAttribute<?>... attrs)
throws IOException
Creates a new directory in the specified directory, using the given prefix to generate its name. The resulting Path is associated with the same FileSystem as the given directory.
The details as to how the name of the directory is constructed is implementation dependent and therefore not specified. Where possible the prefix is used to construct candidate names.
This effectively resolves the embarrassingly ancient bug report in the Sun bug tracker which asked for just such a function.
This is the source code to the Guava library's Files.createTempDir(). It's nowhere as complex as you might think:
public static File createTempDir() {
File baseDir = new File(System.getProperty("java.io.tmpdir"));
String baseName = System.currentTimeMillis() + "-";
for (int counter = 0; counter < TEMP_DIR_ATTEMPTS; counter++) {
File tempDir = new File(baseDir, baseName + counter);
if (tempDir.mkdir()) {
return tempDir;
}
}
throw new IllegalStateException("Failed to create directory within "
+ TEMP_DIR_ATTEMPTS + " attempts (tried "
+ baseName + "0 to " + baseName + (TEMP_DIR_ATTEMPTS - 1) + ')');
}
By default:
private static final int TEMP_DIR_ATTEMPTS = 10000;
See here
Do not use deleteOnExit() even if you explicitly delete it later.
Google 'deleteonexit is evil' for more info, but the gist of the problem is:
deleteOnExit() only deletes for normal JVM shutdowns, not crashes or killing the JVM process.
deleteOnExit() only deletes on JVM shutdown - not good for long running server processes because:
The most evil of all - deleteOnExit() consumes memory for each temp file entry. If your process is running for months, or creates a lot of temp files in a short time, you consume memory and never release it until the JVM shuts down.
As of Java 1.7 createTempDirectory(prefix, attrs) and createTempDirectory(dir, prefix, attrs) are included in java.nio.file.Files
Example:
File tempDir = Files.createTempDirectory("foobar").toFile();
This is what I decided to do for my own code:
/**
* Create a new temporary directory. Use something like
* {#link #recursiveDelete(File)} to clean this directory up since it isn't
* deleted automatically
* #return the new directory
* #throws IOException if there is an error creating the temporary directory
*/
public static File createTempDir() throws IOException
{
final File sysTempDir = new File(System.getProperty("java.io.tmpdir"));
File newTempDir;
final int maxAttempts = 9;
int attemptCount = 0;
do
{
attemptCount++;
if(attemptCount > maxAttempts)
{
throw new IOException(
"The highly improbable has occurred! Failed to " +
"create a unique temporary directory after " +
maxAttempts + " attempts.");
}
String dirName = UUID.randomUUID().toString();
newTempDir = new File(sysTempDir, dirName);
} while(newTempDir.exists());
if(newTempDir.mkdirs())
{
return newTempDir;
}
else
{
throw new IOException(
"Failed to create temp dir named " +
newTempDir.getAbsolutePath());
}
}
/**
* Recursively delete file or directory
* #param fileOrDir
* the file or dir to delete
* #return
* true iff all files are successfully deleted
*/
public static boolean recursiveDelete(File fileOrDir)
{
if(fileOrDir.isDirectory())
{
// recursively delete contents
for(File innerFile: fileOrDir.listFiles())
{
if(!FileUtilities.recursiveDelete(innerFile))
{
return false;
}
}
}
return fileOrDir.delete();
}
Well, "createTempFile" actually creates the file. So why not just delete it first, and then do the mkdir on it?
This code should work reasonably well:
public static File createTempDir() {
final String baseTempPath = System.getProperty("java.io.tmpdir");
Random rand = new Random();
int randomInt = 1 + rand.nextInt();
File tempDir = new File(baseTempPath + File.separator + "tempDir" + randomInt);
if (tempDir.exists() == false) {
tempDir.mkdir();
}
tempDir.deleteOnExit();
return tempDir;
}
As discussed in this RFE and its comments, you could call tempDir.delete() first. Or you could use System.getProperty("java.io.tmpdir") and create a directory there. Either way, you should remember to call tempDir.deleteOnExit(), or the file won't be deleted after you're done.
Just for completion, this is the code from google guava library. It is not my code, but I think it is valueable to show it here in this thread.
/** Maximum loop count when creating temp directories. */
private static final int TEMP_DIR_ATTEMPTS = 10000;
/**
* Atomically creates a new directory somewhere beneath the system's temporary directory (as
* defined by the {#code java.io.tmpdir} system property), and returns its name.
*
* <p>Use this method instead of {#link File#createTempFile(String, String)} when you wish to
* create a directory, not a regular file. A common pitfall is to call {#code createTempFile},
* delete the file and create a directory in its place, but this leads a race condition which can
* be exploited to create security vulnerabilities, especially when executable files are to be
* written into the directory.
*
* <p>This method assumes that the temporary volume is writable, has free inodes and free blocks,
* and that it will not be called thousands of times per second.
*
* #return the newly-created directory
* #throws IllegalStateException if the directory could not be created
*/
public static File createTempDir() {
File baseDir = new File(System.getProperty("java.io.tmpdir"));
String baseName = System.currentTimeMillis() + "-";
for (int counter = 0; counter < TEMP_DIR_ATTEMPTS; counter++) {
File tempDir = new File(baseDir, baseName + counter);
if (tempDir.mkdir()) {
return tempDir;
}
}
throw new IllegalStateException(
"Failed to create directory within "
+ TEMP_DIR_ATTEMPTS
+ " attempts (tried "
+ baseName
+ "0 to "
+ baseName
+ (TEMP_DIR_ATTEMPTS - 1)
+ ')');
}
I got the same problem so this is just another answer for those who are interested, and it's similar to one of the above:
public static final String tempDir = System.getProperty("java.io.tmpdir")+"tmp"+System.nanoTime();
static {
File f = new File(tempDir);
if(!f.exists())
f.mkdir();
}
And for my application, I decided that to add in a option to clear the temp on exit so I added in a shut-down hook:
Runtime.getRuntime().addShutdownHook(new Thread() {
#Override
public void run() {
//stackless deletion
String root = MainWindow.tempDir;
Stack<String> dirStack = new Stack<String>();
dirStack.push(root);
while(!dirStack.empty()) {
String dir = dirStack.pop();
File f = new File(dir);
if(f.listFiles().length==0)
f.delete();
else {
dirStack.push(dir);
for(File ff: f.listFiles()) {
if(ff.isFile())
ff.delete();
else if(ff.isDirectory())
dirStack.push(ff.getPath());
}
}
}
}
});
The method delete all subdirs and files before deleting the temp, without using the callstack (which is totally optional and you could do it with recursion at this point), but I want to be on the safe side.
As you can see in the other answers, no standard approach has arisen.
Hence you already mentioned Apache Commons, I propose the following approach using FileUtils from Apache Commons IO:
/**
* Creates a temporary subdirectory in the standard temporary directory.
* This will be automatically deleted upon exit.
*
* #param prefix
* the prefix used to create the directory, completed by a
* current timestamp. Use for instance your application's name
* #return the directory
*/
public static File createTempDirectory(String prefix) {
final File tmp = new File(FileUtils.getTempDirectory().getAbsolutePath()
+ "/" + prefix + System.currentTimeMillis());
tmp.mkdir();
Runtime.getRuntime().addShutdownHook(new Thread() {
#Override
public void run() {
try {
FileUtils.deleteDirectory(tmp);
} catch (IOException e) {
e.printStackTrace();
}
}
});
return tmp;
}
This is preferred since apache commons the library that comes as closest to the asked "standard" and works with both JDK 7 and older versions. This also returns an "old" File instance (which is stream based) and not a "new" Path instance (which is buffer based and would be the result of JDK7's getTemporaryDirectory() method) -> Therefore it returns what most people need when they want to create a temporary directory.
Try this small example:
Code:
try {
Path tmpDir = Files.createTempDirectory("tmpDir");
System.out.println(tmpDir.toString());
Files.delete(tmpDir);
} catch (IOException e) {
e.printStackTrace();
}
Imports:
java.io.IOException
java.nio.file.Files
java.nio.file.Path
Console output on Windows machine:
C:\Users\userName\AppData\Local\Temp\tmpDir2908538301081367877
Comment:
Files.createTempDirectory generates unique ID atomatically - 2908538301081367877.
Note:
Read the following for deleting directories recursively:
Delete directories recursively in Java
I like the multiple attempts at creating a unique name but even this solution does not rule out a race condition. Another process can slip in after the test for exists() and the if(newTempDir.mkdirs()) method invocation. I have no idea how to completely make this safe without resorting to native code, which I presume is what's buried inside File.createTempFile().
Before Java 7 you could also:
File folder = File.createTempFile("testFileUtils", ""); // no suffix
folder.delete();
folder.mkdirs();
folder.deleteOnExit();
Using File#createTempFile and delete to create a unique name for the directory seems ok. You should add a ShutdownHook to delete the directory (recursively) on JVM shutdown.