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? Thank you Christian -- http://mail.python.org/mailman/listinfo/python-list
