[Zope3-dev] how-to fool XML-RPC publisher

[Adam Groszer]

Hello Stephan,

The most simple and 'brutal' way that I found is this:

def _stringify(string):
    return string

import xmlrpclib
xmlrpclib._stringify = _stringify

What do you think about it?

Sunday, September 11, 2005, 4:26:34 PM, you wrote:

> On Wednesday 31 August 2005 09:23, Groszer Adam wrote:
>> Sorry, I'm a newbie regarding that. As I checked there is a
>> zope.publisher.xmlrpc.premarshal_dispatch_table dict, but this is for
>> the response. The request is handled 'directly' by xmlrpclib
>> self._args, function = xmlrpclib.loads(self._body_instream.read())
>> Can you please give a hint where to look?

> I don't know much about all of this either. You definitely want to register
> this hook with xmlrpclib, so the best place to start would be the Python
> docs. If they don't do what you want, you want to write a proposal to make
> Zope 3 more flexible.

> Regards,
> Stephan



-- 
Best regards,
 Adam                            mailto:adamg at fw.hu
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