Get name of executable jar from within main() method [duplicate]

Question

This question already has an answer here:

• How to get the path of a running JAR file? 28 answers

I've created an executable jar and using commons-cli to give the user the ability to specify command line parameters when he launches the client. Everything works fine. However, when I print the usage statement for the jar, I would like to show the following:

usage: java -jar myprog.jar <options> <file>
--help Display the help message
--debug Enable debugging
....

Printing of all the options is easily done with commons-cli. However, the "usage" line is the head scratcher. I cannot seem to figure out a way to get the "myprog.jar" name from the args[] that are passed to the application.

Is there any easy way of doing this? I could use a pretty convoluted method to back trace from my class' classloader and figure out if it is contained within a jar, but that seems like a fairly ugly answer to what should be a pretty simple question.

private String getPath(Class cls) {
    String cn = cls.getName();
    String rn = cn.replace('.', '/') + ".class";
    String path =
            getClass().getClassLoader().getResource(rn).getPath();
    int ix = path.indexOf("!");
    if(ix >= 0) {
        return path.substring(0, ix);
    } else {
        return path;
    }
}

Answer 1

Here you go:

new java.io.File(SomeClassInYourJar.class.getProtectionDomain()
  .getCodeSource()
  .getLocation()
  .getPath())
.getName()

Edit: I saw your comment about getSourceCode API. Well, this is probably the best you can do in Java. About getCodeSource() returning null, I think it mainly happens on classes in java.lang.* and other special classes for which the source location is "hidden". Should work for your own classes though.

Answer 2

You should in any case add a .toURI() before you call getPath(). Thats because of some quirks in java's URL implementation - see how to encode URL to avoid special characters in java for details.