From: guido at eric.cnri.reston.va.us (Guido van Rossum)
Date: 22 Apr 1999 08:25:30 -0400
Subject: Directory of current file
References: <Pine.WNT.4.05.9904212159220.191-100000@david.ski.org> <371F0BD2.7D40A957@rubic.com> <00ee01be8cba$6e98dcb0$f29b12c2@pythonware.com>
Message-ID: <5lvheosuyt.fsf@eric.cnri.reston.va.us>
X-UID: 550                                                  

"Fredrik Lundh" <fredrik at pythonware.com> writes:

> how about a combination?
> 
> import sys, os
> if __name__ == '__main__':
>     _thisDir = sys.argv[0]
> else:
>     _thisDir = sys.modules[__name__].__file__

Eh, what's wrong with simply using __file__?  It's a global in your
own module, remember!

> _thisDir = os.path.split(_thisDir)[0]

One more refinement: instead of os.path.split(x)[0], use
os.path.dirname(x).

The whole thing could become a one-liner:

_thisDir = os.path.dirname(__name__ == '__main__' and sys.argv[0] or __file__)

You could also use the little-known fact that sys.path[0] is the
script's directory; or the empty string if it's the current directory:

if __name__ == '__main__':
    _thisDir = sys.path[0] or os.curdir
else:
    _thisDir = os.path.dirname(__file__)

--Guido van Rossum (home page: http://www.python.org/~guido/)




