Thanks, but I get this error when I try this.
UnicodeEncodeError: 'ascii' codec can't encode character u'\xe8' in position 12: ordinal not in range(128) I had encountered it with the askdirectory method as well. Is there an easy way to bypass this? Thanks again Tim Daneliuk wrote: > [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? > > > > Thank you > > > > Christian > > > > How about: > > print file.name > > > > -- > ---------------------------------------------------------------------------- > Tim Daneliuk [EMAIL PROTECTED] > PGP Key: http://www.tundraware.com/PGP/ -- http://mail.python.org/mailman/listinfo/python-list
