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How to get the path of a running JAR file?

Posted by: admin November 2, 2017 Leave a comment

Questions:

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.

Answers:
return new File(MyClass.class.getProtectionDomain().getCodeSource().getLocation().toURI().getPath());

Obviously, this will do odd things if your class was loaded from a non-file location.

Questions:
Answers:

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.

Questions:
Answers:

To obtain the File for a given Class, there are two steps:

  1. Convert the Class to a URL
  2. 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.

  1. URL url = Bar.class.getProtectionDomain().getCodeSource().getLocation();

  2. 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:

Questions:
Answers:

You can also use:

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

Questions:
Answers:

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.

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Answers:

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());

Questions:
Answers:

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();
}

Questions:
Answers:

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 "";
    }

Questions:
Answers:

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.

Questions:
Answers:

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"));

Questions:
Answers:

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).

Questions:
Answers:

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.?

Questions:
Answers:
String path = getClass().getResource("").getPath();

The path always refers to the resource within the jar file.

Questions:
Answers:

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.

Questions:
Answers:

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.

Questions:
Answers:

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();

Questions:
Answers:

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();

Questions:
Answers:
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

Questions:
Answers:

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");
}

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Answers:

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");

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Answers:

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;
    }

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Answers:

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;
  }

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Answers:

Ignore backup lad answer, it may look ok sometimes but has several problems:

here both should be +1 not -1:

name = name.substring(name.lastIndexOf(':')-1, name.lastIndexOf('/')+1).replace('%', ' ');

Very dangerous because is not immediately evident if the path has no white spaces, but replacing just the “%” will leave you with a bunch of 20 in each white space:

name = name.substring(name.lastIndexOf(':')-1, name.lastIndexOf('/')+1).replace('%', ' ');

There are better ways than that loop for the white spaces.

Also it will cause problems at debugging time.

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Answers:

This method, called from code in the archive, returns the folder where the .jar file is. It should work in either Windows or Unix.


  private String getJarFolder() {
    String name = this.getClass().getName().replace('.', '/');
    String s = this.getClass().getResource("/" + name + ".class").toString();
    s = s.replace('/', File.separatorChar);
    s = s.substring(0, s.indexOf(".jar")+4);
    s = s.substring(s.lastIndexOf(':')-1);
    return s.substring(0, s.lastIndexOf(File.separatorChar)+1);
  } 

Derived from code at: Determine if running from JAR

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Answers:

I write in Java 7, and test in Windows 7 with Oracle’s runtime, and Ubuntu with the open source runtime. This works perfect for those systems:

The path for the parent directory of any running jar file (assuming the class calling this code is a direct child of the jar archive itself):

try {
    fooDir = new File(this.getClass().getClassLoader().getResource("").toURI());
} catch (URISyntaxException e) {
    //may be sloppy, but don't really need anything here
}
fooDirPath = fooDir.toString(); // converts abstract (absolute) path to a String

So, the path of foo.jar would be:

fooPath = fooDirPath + File.separator + "foo.jar";

Again, this wasn’t tested on any Mac or older Windows

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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, "");
}

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Answers:

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.

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Or you can pass throw the current Thread like this :

String myPath = Thread.currentThread().getContextClassLoader().getResource("filename").getPath();

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This one liner works for folders containing spaces or special characters (like ç or õ). The original question asks for the absolute path (working dir), without the JAR file itself. Tested in here with Java7 on Windows7:

String workingDir = System.getProperty("user.dir");

Reference: http://www.mkyong.com/java/how-to-get-the-current-working-directory-in-java/