problem with xmlrpc server

Question

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 ?

Community · Accepted Answer · 2017-05-23 09:58:46Z

2

Use a timeout:

Set timeout for xmlrpclib.ServerProxy

EDIT

The answer linked to here is not compatible with Python 2.7. Here is modified code that works (tested on W7/ActivePython 2.7):

import xmlrpclib
import httplib

class TimeoutHTTPConnection(httplib.HTTPConnection):

    def __init__(self,host,timeout=10):
        httplib.HTTPConnection.__init__(self,host,timeout=timeout)
        self.set_debuglevel(99)
        #self.sock.settimeout(timeout)

"""
class TimeoutHTTP(httplib.HTTP):
    _connection_class = TimeoutHTTPConnection
    def set_timeout(self, timeout):
        self._conn.timeout = timeout
"""

class TimeoutTransport(xmlrpclib.Transport):
    def __init__(self, timeout=10, *l, **kw):
        xmlrpclib.Transport.__init__(self,*l,**kw)
        self.timeout=timeout

    def make_connection(self, host):
        conn = TimeoutHTTPConnection(host,self.timeout)
        return conn

class TimeoutServerProxy(xmlrpclib.ServerProxy):
    def __init__(self,uri,timeout=10,*l,**kw):
        kw['transport']=TimeoutTransport(timeout=timeout, use_datetime=kw.get('use_datetime',0))
        xmlrpclib.ServerProxy.__init__(self,uri,*l,**kw)

if __name__ == "__main__":
    s=TimeoutServerProxy('http://127.0.0.1:8888',timeout=2)
    print s.test()

edited May 23, 2017 at 9:58

CommunityBot

11 silver badge

answered Apr 14, 2011 at 15:13

AJ.

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1 Comment

Bdfy Over a year ago

this don't work wish python 2.7 <pre> response = h.getresponse(buffering=True) AttributeError: TimeoutHTTP instance has no attribute 'getresponse' </pre>

khachik · Accepted Answer · 2011-04-14 15:27:34Z

0

Make your Test instance daemonic, to quit when the main thread quits:

ts = Test()
ts.setDaemon(True)
ts.start()

The question is why you need to register a thread as an XML-RPC handler.

answered Apr 14, 2011 at 15:27

khachik

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