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.
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Answer
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.