How to get the current running module path/name

Question

I\'ve searched and this seems to be a simple question without a simple answer.

I have the file a/b/c.py which would be called with python -m a.b.c

Answer 1

This works for me:

__loader__.fullname

Also if I do python -m b.c from a\ I get 'b.c' as expected.

Not entirely sure what the __loader__ attribute is so let me know if this is no good.

edit: It comes from PEP 302: http://www.python.org/dev/peps/pep-0302/

Interesting snippets from the link:

The load_module() method has a few responsibilities that it must fulfill before it runs any code:

...

• It should add an __loader__ attribute to the module, set to the loader object. This is mostly for introspection, but can be used for importer-specific extras, for example getting data associated with an importer.

So it looks like it should work fine in all cases.

Answer 2

The only way is to do path manipulation with os.getcwd(), os.path, file and whatnot, as you mentioned.

Actually, it could be a good patch to implement for optparse / argparse (which currently replace "%prog" in the usage string with os.path.basename(sys.argv[0]) -- you are using optparse, right? -- ), i.e. another special string like %module.