Using SimpleXMLRPCSer ver in a Windows Service

[Rudy Schockaert]

After some Googling I found a post of someone who wanted to do exactly
as what I want to do now.
There is however a problem in his code that makes the service fails
after the first connection. I slightly modified his code and now I can
run the service longer before I run into trouble.
I then tried making the SimpleXMLRPCSer ver multi-threaded, hoping the
problem would disappear, but no avail.
The code is as follows:
The commented part in the while loop is from the original code.
<CODE>
## XML-RPC Service
import sys
import win32serviceuti l
import win32service
import win32event
import win32evtlogutil
import win32file
import servicemanager
import SimpleXMLRPCSer ver
import SocketServer
import select

class OBJECT:
def hello(self, text):
return "Hello World (%s)" % text

class ThreadedSimpleX MLRPCServer(Soc ketServer.Threa dingMixIn,
SimpleXMLRPCSer ver.SimpleXMLRP CServer): pass

class XMLRPCSERVICE(w in32serviceutil .ServiceFramewo rk):
_svc_name_ = "XMLRPCSERV ICE"
_svc_display_na me_ = "XMLRPCSERV ICE"
_svc_descriptio n_ = "XMLRPCSERV ICE"

def __init__(self, args):
win32evtlogutil .AddSourceToReg istry(self._svc _display_name_,
sys.executable, "Applicatio n")
win32serviceuti l.ServiceFramew ork.__init__(se lf, args)

self.hWaitStop = win32event.Crea teEvent(None, 0, 0, None)
self.hSockEvent = win32event.Crea teEvent(None, 0, 0, None)
self.stop_reque sted = 0

def SvcStop(self):
self.ReportServ iceStatus(win32 service.SERVICE _STOP_PENDING)
self.stop_reque sted = 1
win32event.SetE vent(self.hWait Stop)

def SvcDoRun(self):
## Write a started event
servicemanager. LogMsg(
servicemanager. EVENTLOG_INFORM ATION_TYPE,
servicemanager. PYS_SERVICE_STA RTED,
(self._svc_name _, ' (%s)' % self._svc_name_ ))

server = ThreadedSimpleX MLRPCServer(("" , 8080))
object = OBJECT()
server.register _instance(objec t)
self.socket = server.socket

while 1:
r, w, x = select.select([self.socket],[],[],10)
if r == [self.socket]:
server.handle_r equest()
if self.stop_reque sted:
self.socket.clo se()
break
#win32file.WSAE ventSelect(serv er,
self.hSockEvent ,win32file.FD_A CCEPT)
#rc =
win32event.Wait ForMultipleObje cts((self.hWait Stop,self.hSock Event), 0,
win32event.INFI NITE)
#if rc == win32event.WAIT _OBJECT_0:
# break
#else:
# server.handle_r equest()
# win32file.WSAEv entSelect(serve r,self.hSockEve nt, 0)
# #server.serve_f orever() ## Works, but breaks the

## Write a stopped event
win32evtlogutil .ReportEvent(se lf._svc_name_,
servicemanager. PYS_SERVICE_STO PPED,0,
servicemanager. EVENTLOG_INFORM ATION_TYPE,
(self._svc_name _,""))

if __name__ == '__main__':
win32serviceuti l.HandleCommand Line(XMLRPCSERV ICE)
</CODE>

I tested with the following:

<CODE>

import xmlrpclib
import time

server = xmlrpclib.Serve rProxy("http://localhost:8080" )
for i in range(100):
print server.hello("% d" % i)
time.sleep(1)

</CODE>

The loop ends with the following error:

<OUTPUT>
Hello World (0)
....
Hello World (44)
Traceback (most recent call last):
File "C:\Python24\Li b\site-packages\python win\pywin\frame work\scriptutil s.py",
line 310, in RunScript
exec codeObject in __main__.__dict __
File "C:\DATA\TestSo ap.py", line 6, in ?
print server.hello("% d" % i)
File "C:\Python24\li b\xmlrpclib.py" , line 1096, in __call__
return self.__send(sel f.__name, args)
File "C:\Python24\li b\xmlrpclib.py" , line 1383, in __request
verbose=self.__ verbose
File "C:\Python24\li b\xmlrpclib.py" , line 1137, in request
headers
ProtocolError: <ProtocolErro r for localhost:8080/RPC2: -1 >
</OUTPUT>

Can someone help me in creating a windows service that allows me to
handle XMLRPC request?

Thanks in advance,

Rudy Schockaert