[Tutor] finding path to resource files in a GUI

Christian Meesters meesters at uni-mainz.de
Thu Jun 9 12:43:42 CEST 2005

On 9 Jun 2005, at 12:00, Michael Lange wrote:

> Hi Christian,
> try
>     self.cwd = os.path.abspath(sys.path[0])
> sys.path[0] is the directory of your main program file, with 
> os.path.abspath() you can
> get rid of the "../../" stuff at the beginning of the path if the 
> program is called from a link.
> I hope this helps

It does! Actually I tried abspath, but didn't think of this very 
combination with sys.path ...

Thanks a lot!

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