How to get the path of a running JAR file?

Answer 1

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.

Answer 2

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);

Answer 3

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.

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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:

Answer 4

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.

Answer 5

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

Answer 6

You can also use:

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