[BangPypers] XML RPC Problem....
uzmanajmal at gmail.com
Sat Sep 13 12:06:11 CEST 2008
Where exactly should i call ServerProxy? Following is the code from my
t = SecureTransport()
server = xmlrpclib.Server('http://localhost:8000/',transport=t)
Note: Full code for client is here at http://privatepaste.com/b56oS1Xa7P
and following is my server code
#server = ServerProxy("http://betty.userland.com")
server = SimpleXMLRPCServer.SimpleXMLRPCServer(("localhost", 8000))
On Sat, Sep 13, 2008 at 8:44 AM, Fredrik Lundh <fredrik at pythonware.com>wrote:
> Usman Ajmal wrote:
> Please explain the arguments of send_request. What exactly are the
>> connection, handler and request_body? It will be really helpful if you give
>> an example of how do i call send_request
> you don't call send_request. you should pass the SecureTransport instance
> as an argument to the ServerProxy, which will then use it to talk to the
> server. see the "custom transport" example in the library reference that I
> pointed you to.
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