The following should work: os.path.split( os.path.realpath( sys.argv[0] ) )[0] Cheers,
pieter -----Original Message----- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Gabriel Cooper Sent: 24 January 2005 16:40 To: python-list@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? -- http://mail.python.org/mailman/listinfo/python-list -- http://mail.python.org/mailman/listinfo/python-list
