Paul McGuire wrote:
4. filename=r[7].split('/')[-1] is not terribly portable. See if there is a
standard module for parsing filespecs (I'll bet there is).
Indeed there is -- os.path. In particular, os.path.basename() seems to do exactly that snippet is intending, in a much more robust (and readable) fashion.
Jeff Shannon Technician/Programmer Credit International
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