How to get current class name including package name in Java?

Question

I'm working on a project and one requirement is if the 2nd argument for the main method starts with “/” (for linux) it should consider it as an absolute path (not a problem), but if it doesn't start with “/”, it should get the current working path of the class and append to it the given argument.

I can get the class name in several ways: System.getProperty("java.class.path"), new File(".") and getCanonicalPath(), and so on...

The problem is, this only gives me the directory in which the packages are stored - i.e. if I have a class stored in ".../project/this/is/package/name", it would only give me "/project/" and ignores the package name where the actual .class files lives.

Any suggestions?

EDIT: Here's the explanation, taken from the exercise description

sourcedir can be either absolute (starting with “/”) or relative to where we run the program from

sourcedir is a given argument for the main method. how can I find that path?

Answer 1

Use this.getClass().getCanonicalName() to get the full class name.

Note that a package / class name ("a.b.C") is different from the path of the .class files (a/b/C.class), and that using the package name / class name to derive a path is typically bad practice. Sets of class files / packages can be in multiple different class paths, which can be directories or jar files.

Answer 2

There is a class, Class, that can do this:

Class c = Class.forName("MyClass"); // if you want to specify a class
Class c = this.getClass();          // if you want to use the current class

System.out.println("Package: "+c.getPackage()+"\nClass: "+c.getSimpleName()+"\nFull Identifier: "+c.getName());

If c represented the class MyClass in the package mypackage, the above code would print:

Package: mypackage
Class: MyClass
Full Identifier: mypackage.MyClass

You can take this information and modify it for whatever you need, or go check the API for more information.