SimpleXMLRPCServer and client IP address

Jeremy Monnet jmonnet at
Wed Jun 28 17:05:05 CEST 2006


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 :-)


from liste_films import *
import SimpleXMLRPCServer

def isOpen():
   # Here I want to get the clien IP address
   if CheckPerms(IP):
       return True
       return False

if __name__ == '__main__':
   server = SimpleXMLRPCServer.SimpleXMLRPCServer(("localhost", 8000))

#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
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

I'm using python2.3 on linux (debian sid).

Thanks for any help !

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.

More information about the Python-list mailing list