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Re: [Gnumed-devel] xmlrpc performance
From: |
Horst Herb |
Subject: |
Re: [Gnumed-devel] xmlrpc performance |
Date: |
Fri, 28 Feb 2003 10:39:00 +1100 |
User-agent: |
Mozilla/5.0 (Windows; U; Win98; en-US; rv:1.3b) Gecko/20030210 |
Horst Herb wrote:
if no_xmlrpc
import myserver
def connect():
return xmlrpclib.Server('http://localhost:8000')
if no_xmlrpc:
server = myserver.PersonDB()
else:
server = connect()
server.GetPerson('1')
Sorry, must have been sleeping while typing this crappy design.
Should be:
def connect(xmlrpc_host=None):
if xmlrpc_host is not None:
return xmlrpclib.Server(xmlrpc_host)
else:
import my_server_module
return my_server_module.my_service_class()
and then, anything in the rest of the client module will be entirely
transparent; no "if's" needed at all.
Any service module, if run as "__main__", should run as XMLRPC server
and use sys.argv[1] as port argument
If imported, it should have a function "runXMLRPCserver(port)" to start
the server programmatically if needed
Of course, one would polish up connnect() a bit to implement lazy
/cached connections/objects
Horst