Hi,
> This function would take two paths: A and B and give
> the relation between them. Here are a few of examples.
>
> os.path.diff("/A/C/D/", "/A/D/F/")
> ==> "../../D/F"
>
> os.path.diff("/A/", "/A/B/C/")
> ==> "B/C"
>
> os.path.diff("/A/B/C/", "/A/")
> ==> "../.."
I'm not sure whether something like this is generally non-trivial.
Suppose you have the following directory structure:
/home -> usr/home
/usr
/usr/home
What does os.path.diff("/home/", "/usr/") yield? "../usr/", I would
presume? But that's obviously wrong:
>>> import os
>>> os.stat("/home/../usr")
Traceback (most recent call last):
File "<stdin>", line 1, in ?
OSError: [Errno 2] No such file or directory: '/home/../usr'
I've not thought about this long enough to say if there's a trivial
solution to this, though :)
Bye,
Matthias A. Benkard
________________________________________________________________________
Matthias Andreas Benkard, Anarchokommunist und Pythonprogrammierer
Persönliche Website: http://www.mulk.de.vu/
Persönlicher Weblog: http://www.kompottkin.de.vu/
Flames bitte nach /dev/null schicken.
signature.asc
Description: This is a digitally signed message part
_______________________________________________ Python-Dev mailing list Python-Dev@python.org http://mail.python.org/mailman/listinfo/python-dev Unsubscribe: http://mail.python.org/mailman/options/python-dev/archive%40mail-archive.com
