On Wed, 08 Nov 2006 20:01:08 +0100, <[EMAIL PROTECTED]> wrote: > Hello, > I am working on a school project that requires me to get the path of a > filename for future treatment. > I've tried getting a file with tkFileDialog.askopenfile. > > > ******************************************** > import tkFileDialog > file = tkFileDialog.askopenfile() > print file > ******************************************** > > > It prints the opened files stuff, but I just can not find how to get > that path as a string. I've searched around google and the present > group, and found no documentation on the file class used with > tkFileDialog. Does someone have a solution for that?
Use tkFileDialog.askopenfilename? HTH -- python -c "print ''.join([chr(154 - ord(c)) for c in 'U(17zX(%,5.zmz5(17l8(%,5.Z*(93-965$l7+-'])" -- http://mail.python.org/mailman/listinfo/python-list
