Generating Filenames from Feeds

Joel Goldstick joel.goldstick at gmail.com
Thu Mar 14 17:05:05 CET 2013


On Thu, Mar 14, 2013 at 11:38 AM, Chuck <galois271 at gmail.com> wrote:

> HI all,
>
> I am trying to write a podcast catcher for fun, and I am trying to come up
> with a way to generate a destination filename to use in the function
> urlretrieve(url, destination).  I  would like the destination filename to
> end in a .mp3 extension.
>
> My first attempts were parsing out the <pubdate> and stripping the
> whitespace characters, and joining with os.path.join.  I haven't been able
> to make that work for some reason.


The reason is apparently a  syntax error.

Whenever I put the .mp3 in the os.path.join I get syntax errors.  I am
> wondering if there is a better way?
>

Yes, don't write code with syntax errors!

>
> I was doing something like
> os.path.join('C:\\Users\\Me\\Music\\Podcasts\\', pubdate.mp3), where
> pubdate has been parsed and stripped of whitespace.  I keep getting an
> error around the .mp3.
>
> Any ideas?
>

Seriously, if you don't post a minimal code example that shows the problem
and with a full traceback you are asking strangers to do magic tricks for
your pleasure.

>
> Thanks!!
> Chuck
> --
> http://mail.python.org/mailman/listinfo/python-list
>



-- 
Joel Goldstick
http://joelgoldstick.com
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