Get location of JAR file

Question

I'm trying to get the location a Runnable JAR file is run from. I tried doing

try {
    String path = new java.io.File(".").getCanonicalPath();
} catch (IOException e) {
    e.printStackTrace();
}

But that returns:

C:\Users\Kevin\Desktop/server/Server

while the JAR file is located at

C:\Users\Kevin\Desktop

I also tried doing

return new file(Server.class.getProtectionDomain().getCodeSource().getLocation().getPath());

But that returns:

C:\Users\Kevin\Desktop\server.jar/server/Server

So basicly I want the path of the JAR file without the filename and not the ClassPath.

Any way of doing this?

This might be helpful, pervious discussion. http://stackoverflow.com/questions/320542/how-to-get-the-path-of-a-running-jar-file //Henrik — Hiny, Mar 12 '13 at 11:21
The specification is incomplete. What should be done if you're *not* running from a JAR? What if the JAR lives in some Network? — Ingo, Mar 12 '13 at 11:23
Well the thing is that i'm trying to create a directory containing files and that has to be on my desktop or wherever i'm running the JAR from. If I run it in debug mode within Eclipse it does create files in the bin/server folder. — Snakybo, Mar 12 '13 at 11:27

BackSlash · Accepted Answer · 2017-02-16T16:00:28.930

29

Update

Since it doesn't work in certain test-cases, I'll update the answer.

The correct way to do this should be with ClassLoader:

File jarDir = new File(ClassLoader.getSystemClassLoader().getResource(".").getPath());
System.out.println(jarDir.getAbsolutePath());

Tested on various classpaths, the output was correct.

Old answer

This should work

File f = new File(System.getProperty("java.class.path"));
File dir = f.getAbsoluteFile().getParentFile();
String path = dir.toString();

It works for me, my program is in:

C:\Users\User01\Documents\app1\dist\JavaApplication1.jar

And it returns

C:\Users\User01\Documents\app1\dist

edited Feb 16 '17 at 16:00

answered Mar 12 '13 at 11:32

BackSlash

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5

This works only so long as you classpath contains just one entry. – Ingo Mar 12 '13 at 11:40
@BackSlash have you ever tried the above code on linux/unix system does this work perfectly? – Deepak Bhatia Jan 04 '14 at 07:57
@Ingo "classpath contains just one entry" what it actually means...?? – Deepak Bhatia Jan 04 '14 at 08:03
5

@dbw The classpath can be a list of jar files, zip files and directories, separated by ':' in Unix or ';' in Windows. Clearly `new File("foo.jar:bar/:baz.jar")` wouldn't work then. – Ingo Jan 04 '14 at 13:19
@Ingo but when `System.getProperty("java.class.path")` will return list of jar files as you mentioned – Deepak Bhatia Jan 04 '14 at 13:22
1

When you set it that way, for example running `java -cp foo.jar:bar/:baz.jar .....` – Ingo Jan 04 '14 at 13:25
@taha It works for me. Maybe you could put up an example demonstrating the issue so that I can take a look. – BackSlash Feb 04 '19 at 09:27
it works in instance methods but not in static main method – taha Jun 18 '19 at 05:51
3

getResource(".") returns null in my case – Nikita Aug 25 '19 at 10:05
doesn't work for me. getResource(".") returns null. – moonlightcheese Jun 23 '20 at 18:29

score 20 · Answer 2 · answered Jan 06 '14 at 15:34

Herewith my version of computing the jar directory

/**
 * Compute the absolute file path to the jar file.
 * The framework is based on http://stackoverflow.com/a/12733172/1614775
 * But that gets it right for only one of the four cases.
 * 
 * @param aclass A class residing in the required jar.
 * 
 * @return A File object for the directory in which the jar file resides.
 * During testing with NetBeans, the result is ./build/classes/,
 * which is the directory containing what will be in the jar.
 */
public static File getJarDir(Class aclass) {
    URL url;
    String extURL;      //  url.toExternalForm();

    // get an url
    try {
        url = aclass.getProtectionDomain().getCodeSource().getLocation();
          // url is in one of two forms
          //        ./build/classes/   NetBeans test
          //        jardir/JarName.jar  froma jar
    } catch (SecurityException ex) {
        url = aclass.getResource(aclass.getSimpleName() + ".class");
        // url is in one of two forms, both ending "/com/physpics/tools/ui/PropNode.class"
        //          file:/U:/Fred/java/Tools/UI/build/classes
        //          jar:file:/U:/Fred/java/Tools/UI/dist/UI.jar!
    }

    // convert to external form
    extURL = url.toExternalForm();

    // prune for various cases
    if (extURL.endsWith(".jar"))   // from getCodeSource
        extURL = extURL.substring(0, extURL.lastIndexOf("/"));
    else {  // from getResource
        String suffix = "/"+(aclass.getName()).replace(".", "/")+".class";
        extURL = extURL.replace(suffix, "");
        if (extURL.startsWith("jar:") && extURL.endsWith(".jar!"))
            extURL = extURL.substring(4, extURL.lastIndexOf("/"));
    }

    // convert back to url
    try {
        url = new URL(extURL);
    } catch (MalformedURLException mux) {
        // leave url unchanged; probably does not happen
    }

    // convert url to File
    try {
        return new File(url.toURI());
    } catch(URISyntaxException ex) {
        return new File(url.getPath());
    }
}

score 2 · Answer 3 · answered Aug 30 '16 at 13:31

2

I had to mess around a lot before I finally found a working 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();

answered Aug 30 '16 at 13:31

Jelle den Burger

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`.getParent()` is necessary for me to get the path for Runnable JAR. – Northnroro Feb 18 '20 at 11:58

score 1 · Answer 4 · answered Mar 12 '13 at 11:28

1

If you know

file(Server.class.getProtectionDomain().getCodeSource().getLocation().getPath());

returns

C:\Users\Kevin\Desktop\server.jar/server/Server

And you know your Jar name is server.jar or for matter any .jar file, your intention is to get C:\Users\Kevin\Desktop , straight forward way is to do string manipulation.

With the retrieved output, tokenize the string based on File.separator and construct the path (by concatenating the strings with File.separator in between) until you get a token which contains .jar

answered Mar 12 '13 at 11:28

Vinod Jayachandran

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Yes but what if the users decides to change the name of the JAR? – Snakybo Mar 12 '13 at 11:29
You should not search with hard coded jar file name, ideally after retrieving the token, you should check if that token.contains(.jar), this will work what ever may be you jar file name – Vinod Jayachandran Mar 12 '13 at 11:31
No it won't, for example if it is renamed ZIP – Ingo Mar 12 '13 at 11:38
It's a simple String manipulation, you need to handle all the possible file extensions for your requirement. Generally it's enough if you just handle jar & zip – Vinod Jayachandran Mar 12 '13 at 11:43

Get location of JAR file

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