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pythonpython-2.6

How to get the current running module path/name

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. I would like to obtain the value a.b.c in the module level.

USAGE = u'''\
Usage:
    python -m %s -h
''' % (what_do_i_put_here,)

So when I receive the -h option, I display the USAGE without the need to actually write down the actual value in each and every script.

Do I really need to go through inspect to get the desired value?

Thanks.

EDIT: As said, there are answers (I've searched), but not simple answers. Either use inspect, use of traceback, or manipulate __file__ and __package__ and do some substring to get the answer. But nothing as simple as if I had a class in the module, I could just use myClass.__module__ and I would get the answer I want. The use of __name__ is (unfortunately) useless as it's always "__main__".

Also, this is in python 2.6 and I cannot use any other versions.

Solution

  • 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.