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

Question

This question already has answers here:

How to get the path of a running JAR file? (33 answers)

Closed 10 years ago.

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;
 }
}

• Charles: tried stackoverflow.com/questions/320542/… ?
• Eric B.: I did see that link already, but it looked as though it was almost a "hack" as well. I cannot believe that there is not a more direct solution. Additionally, the API states that "getCodeSource()" may return null, but does not elaborate under which conditions, so I naturally wondered if this was a "fail-safe" method to use.

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.

• Eric B.: Thanks - that's the same thing that @Charles pointed to in the SO link. My biggest concern as I mentioned above is that the api for getCodeSource() indicates that it can be null, but does not elaborate under which conditions.
• Eric B.: The other problem with this is that it does not take into account the fact that the class file may be nested within archives. So I would still need to check for a ! substring. Like I said, it seems a little clunky and am quite surprised that Java does not have a better solution.
• Charles: It's mostly because there is rarely a reason to know the name of the jar file you're running from. The functionality should (ideally) be independent of the name.
• saravana kumar ramasamy: @rodion for me the above is giving classes! as output
• Sridhar Sarnobat: Any idea if this technique will translate to GraalVM AOTC? I'd like to create a multicall binary.

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.