Finding a script's home directory?

[Pieter Claerhout]

The following should work:

os.path.split( os.path.realpath( sys.argv[0] ) )[0]
 
Cheers,


pieter


-----Original Message-----
From: python-list-bounces+pieter.claerhout=creo.com at python.org
[mailto:python-list-bounces+pieter.claerhout=creo.com at python.org] On Behalf
Of Gabriel Cooper
Sent: 24 January 2005 16:40
To: python-list at python.org
Subject: Finding a script's home directory?

In one of my python programs has a data file I need to load. My solution 
was to say:

        if os.path.exists(os.path.join(os.getcwd(), "config.xml")):
            self.cfgfile = os.path.join(os.getcwd(), "config.xml")

Which works fine... as long as you're *in* the script's home directory 
when you run it (as in, run it as: ./startApp.py as opposed to 
./myApp/startApp.py).

If I run it from an alternate directory the program looks for the 
config.xml file in my current directory not the app's home directory. So 
how do I get the script's home directory?
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