Okay, I changed this:
server = SimpleXMLRPCServer.SimpleXMLRPCServer(("localhost", 8000), ....)
for this:
server = SimpleXMLRPCServer.SimpleXMLRPCServer(('', 8000), ....)
Replacing "localhost" with two simple quotes ' makes it work.
Anyone knows the reason for this?
Thank so much.
Jose Carlos
2006/4/12, Jose Carlos Balderas Alberico <[EMAIL PROTECTED]>:
Thank you for the quick reply John...Is there a way to sort this out? Should I specify another address here:server = SimpleXMLRPCServer.SimpleXMLRPCServer(("localhost", 8000), ....)instead of "localhost" ?I'm kind of new to client/server programming, so I'm at a loss here.Thank you very much for your attention.Jose Carlos.
2006/4/12, John Abel <[EMAIL PROTECTED]>:Your server is only listening on 127.0.0.1.
Jose Carlos Balderas Alberico wrote:
> Up till now I've been setting up my server and client in the same
> machine, using the "localhost" address for everything.
> Once I've made it work, I need to move my client application to the
> computer where it'll be run from, and for some reason, I get a
> socket.error: (111, 'connection refused').
>
> The server is listening on port 8000. I've used the line
>
> server = SimpleXMLRPCServer.SimpleXMLRPCServer(("localhost", 8000), ....)
> ##register functions...
> ...
> ...
> server.serve_forever()
>
>
> And the client does the following:
>
> host = "XXXX:8000" (where XXXX is the server's IP address)
> conn = xmlrpclib.connect(host)
> data = "" (requestData is a function previously
> registered in the server)
>
> I've made sure the server is listening on port 8000, since the netstat
> command says it's listening on port 8000.
> I've also pinged the server from the client and viceversa and I get an
> answer. So they can see each other.
>
> I've tried looking in google but couldn't find a solution to this
> problem. Anyone can give me a hand on this?
> Thank you very much.
>
> Jose Carlos.
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