29 Jun
2005
29 Jun
'05
8:18 p.m.
Hi, maybe I haven't searched enough, but I don't find anything that halps me to solve my problem. I've got a serverapplication. The problem is in the following lines: application = service.Application("TwistedServerApp") internet.TCPServer(8800, pb.ServerFactory...) I don't want to start this with a predetermined port. More like: twistd -y TwistedServerApp.tac 8800 Is this somehow possible? sys.argv seems impossible to use. -Stephan