[Twisted-Python] Get IP of connecting client in XMLRPC server
Remy C. Cool
dev-python at smartology.nl
Fri Jun 11 09:34:56 EDT 2004
Hello,
On 18 May 2004, the following message was posted to this mailinglist:
Jp Calderone exarkun at divmod.com wrote:
>Daniel Newton wrote:
>
> I have a simple XML-PRC server similar to the example below:
>
> from twisted.web import xmlrpc, server
>
> class Example(xmlrpc.XMLRPC):
> """An example object to be published."""
>
> def xmlrpc_add(self, a, b):
> """Return sum of arguments."""
> return a + b
def xmlrpc_whatIsMyAddress(self):
return self.transport.getPeer().host
>
> if __name__ == '__main__':
> from twisted.internet import reactor
> r = Example()
> reactor.listenTCP(7080, server.Site(r))
> reactor.run()
>
> I want to be able to get the address of the client that calls the
xmlrpc
> method can anyone help me with this?
This solution didn't work because 'transport' isn't a property of the
Example class.
I'm currently in the process of changing from a customized
SimpleXMLRPCServer to a twisted XMLRPC server solution and I need to
insert the client IP into the attributes passed to the called xmlrpc
method. Anyone who knows the answer and is willing to share the info?
Regards,
Remy
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