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SimpleXMLRPCServer - revisited

Hello all,

I want to be able to determine the IP address of the client making an
XMPRPC call.

I got the tip to use this:
-------------------------------
class RPCServer(SimpleXMLRPCServer):

def _dispatch(self, method, params):
"""Extend dispatch, adding client info to some parameters."""
if method in ({my list of methods I needed client address}):
return SimpleXMLRPCServer._dispatch(self, method,
params+(self.client_address,))
return SimpleXMLRPCServer._dispatch(self, method, params);
-------------------------------

But it complains that it can not find 'client_address'.

It may be relevant to point out that I am using this:

-------------------------------
class AsyncXMLRPCServer(SocketServer.ThreadingMixIn,Simp leXMLRPCServer):
pass
-------------------------------

If anyone has any hints or tips.. Please share them.

--
Kind regards,
Jan Danielsson
Apr 4 '07 #1
0 992

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