Can't connect to SimpleXMLRPCServer. Help needed.

[John Abel]

Using the '' makes it listen on all interfaces.

Jose Carlos Balderas Alberico wrote:
> 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 
> <josecarlos.balderas at gmail.com <mailto:josecarlos.balderas at gmail.com>>:
>
>     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 <jabel at plus.net <mailto:jabel at plus.net>>:
>
>         Your server is only listening on 127.0.0.1 <http://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 = conn.requestData() (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.
>
>
>      
>
>