469,920 Members | 2,457 Online
Bytes | Developer Community
New Post

Home Posts Topics Members FAQ

Post your question to a community of 469,920 developers. It's quick & easy.

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,
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):

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

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

This discussion thread is closed

Replies have been disabled for this discussion.

Similar topics

2 posts views Thread by Marco Aschwanden | last post: by
3 posts views Thread by Maxim Khesin | last post: by
4 posts views Thread by codecraig | last post: by
reply views Thread by Thomas G. Apostolou | last post: by
2 posts views Thread by Laszlo Nagy | last post: by
reply views Thread by Juju | last post: by
3 posts views Thread by Achim Domma | last post: by
9 posts views Thread by Bret | last post: by
By using this site, you agree to our Privacy Policy and Terms of Use.