[Grok-dev] Re: Using POST request

[Uli Fouquet]

Hi there,

Andreas Jung wrote:
> 
> --On 24. März 2008 12:02:21 +0100 Andreas Jung <lists at zopyx.com> wrote:

> Actually I get intercept a POST request using this approach. But:
> 
>  - sending a POST request through 'curl' works
>  - POSTing to the same URL using 'python2.4 setup.py sdist upload -r URL'
>    still fails with a 404
> 
> The transcript recorded using Wireshark is here:
> 
> <http://zopyx.com/tmp/sg.txt>
> 
> So is there some issue with multipart/form-data?

Well, that seems to be setuptools specific. There's a form field::

   ----------------GHSKFJDLGDS7543FJKLFHRE75642756743254
   Content-Disposition: form-data; name=":action"

   file_upload

requiring an action called 'file_upload'. Because normal views are not
named that way, they cannot be found and this results in a 404 - file
not found.

So, your application must provide this 'action'. With views you should
be able to do it like this::

  class FileUpload(grok.View):
      grok.context(eggserver)
      grok.name('file_upload') # Might do the trick...
      def render(self):
          return "Thanks for submitting."

Haven't tested it. Is something similar possible with REST?

Kind regards,

-- 
Uli

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