I run simple example with xmlrpc server and press Ctrl-C on keyboard :).
from SimpleXMLRPCServer import SimpleXMLRPCServer from time import sleep import threading,time class Test(threading.Thread): def __init__(self): threading.Thread.__init__(self) self.test1 = 0 def test(self): return self.test1 def run(self): while(1): time.sleep(1) self.test1 = self.test1 + 1 ts = Test() ts.start() server = SimpleXMLRPCServer(("localhost",8888)) server.register_instance(ts) server.serve_forever()
error after pressing keyboard:
File "/usr/lib/python2.7/SocketServer.py", line 225, in serve_forever r, w, e = select.select([self], [], [], poll_interval) KeyboardInterrupt
Client
from xmlrpclib import ServerProxy r=ServerProxy("http://localhost:8888") print r.test()waiting connect without error or warning. How to break connection in this case ? Maybe this example is not correct ?