[Python-Dev] os.path.diff(path1, path2)

[Matthias Andreas Benkard]

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/
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