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Re: Generating Filenames from Feeds

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On 14/03/2013 15:38, Chuck 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.  Whenever I put the .mp3 in
> the os.path.join I get syntax errors.  I am wondering if there is a
> better way?
>
> 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?
>
The filename referred to by pubdate is a string, and you want to append
an extension, also a string, to it. Therefore:

os.path.join('C:\\Users\\Me\\Music\\Podcasts\\', pubdate + '.mp3')

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Thread

Generating Filenames from Feeds Chuck <galois271@gmail.com> - 2013-03-14 08:38 -0700
  Re: Generating Filenames from Feeds Joel Goldstick <joel.goldstick@gmail.com> - 2013-03-14 12:05 -0400
    Re: Generating Filenames from Feeds Chuck <galois271@gmail.com> - 2013-03-14 11:19 -0700
      Re: Generating Filenames from Feeds Steven D'Aprano <steve+comp.lang.python@pearwood.info> - 2013-03-14 22:58 +0000
    Re: Generating Filenames from Feeds Chuck <galois271@gmail.com> - 2013-03-14 11:19 -0700
  Re: Generating Filenames from Feeds MRAB <python@mrabarnett.plus.com> - 2013-03-14 16:07 +0000

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