On Tue, Jun 16, 2009 at 2:04 AM, Vadim Zaliva <l...@codeminders.com> wrote:
>
> I am trying to write Python handler which will analyze URL path. For
> example I would like it to be accessible
> as:
>
> /feed/1
> /feed/2
>
> So I've put following to app.yaml:
>
> - url: /feed/.*
> script: feed.py
We also need to see the contents of feed.py - the WSGIApplication section.
>
> In my handler I print path:
>
> def get(self):
> logging.debug("FEED with path '%s'" % self.request.path)
> ...
>
> However regex matching does not seems to be working my handler never
> gets invoked and I am getting 404.
>
> If I remove "/.*" part from URL matching expression, my handler is
> invoked OK for "http://localhost:80800/feed"; URL.
"/feed/.*" will match anything starting with "/feed/". "/feed" does not, so
it won't match. You probably want "/feed(?:/.*)?"
-Nick Johnson
>
> I am testing with dev_appserver.py
>
> What I am doing wrong?
>
> Sincerely,
> Vadim
>
> >
>
--
Nick Johnson, App Engine Developer Programs Engineer
Google Ireland Ltd. :: Registered in Dublin, Ireland, Registration Number:
368047
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