Question

I try to retrive a path of a directory where my executable jar file is situated.
That means: I have the following structure:

 --> Application (this is a folder somewhere in my file system)
     --> application.jar (this is my java application
     --> SomeData (folder in the same directory like the application)
         --> some other folders
         ......

When I start my application.jar via command line I want to parse some files inside the SomeData folder.
In https://stackoverflow.com/a/320595/1540630 they already showed how to get the current path of a running jar file but when I execute the statement:

System.out.println(XMLParser.class.getProtectionDomain().getCodeSource().getLocation().getPath());

…I just get the following:

/.../Application/application.jar

But I just want to have:

/.../Application/

Better to say in later steps I need

/.../Application/SomeData/SomeFolder/data.xml

Can someone help me please?

Answer 1

CodeSource.getLocation() gives you a URL, you can then create new URLs relative to that:

URL jarLocation = XMLParser.class.getProtectionDomain().getCodeSource().getLocation();
URL dataXML = new URL(jarLocation, "SomeData/SomeFolder/data.xml");

You can’t simply do new File(...getCodeSource().getLocation().getPath()) as a URL path is not guaranteed to be a valid native file path on all platforms. You’re much safer sticking with URLs and passing the URL directly to your XML parser if you can, but if you really need a java.io.File then you can use an idiom like this to create one from a URL.