Hello, I've started python a few weeks ago, and to now everything went fine with my cookbook and a learning book.
Now, I've tried the SimpleXMLRPCServer, and it worked OK untill I tried to get the client IP address. I have searched a long time the Internet but couldn't find a _simple_ solution :-) #Code from liste_films import * import SimpleXMLRPCServer def isOpen(): # Here I want to get the clien IP address if CheckPerms(IP): return True else: return False if __name__ == '__main__': server = SimpleXMLRPCServer.SimpleXMLRPCServer(("localhost", 8000)) server.register_function(isOpen) server.serve_forever() #end code Tips I've found were : - inherit from requestDispatcher and overload its methods. But that's not that Simple. - use the requestHandler and its method address_string(), but I didn't an easy to understand example - http://mail.python.org/pipermail/python-list/2006-May/340266.html but this thread seems not to have been finished :-( Furthermore, I think I should be able to access the socket object from where I am (at starting of isOpen() ), but I don't know how. "self" and "parent" are not defined, I can access the "server" object, but it says the other end s not connected ("transport endpoint"). I think this SimpleXMLRPCServer was not threaded (because it says in the API : "answer all requests one at a time"), so I don't understand why in the middle of my function the server.socket.getpeername() says it's not connected. I'm using python2.3 on linux (debian sid). Thanks for any help ! Jeremy -- Linux Registered User #317862 Linux From Scratch Registered User #16571 Please do not send me .doc, .xls, .ppt, as I will *NOT* read them. Please send me only open formats, as OpenDocument or pdf. -- http://mail.python.org/mailman/listinfo/python-list