En Mon, 18 Jun 2007 00:25:25 -0300, <[EMAIL PROTECTED]> escribió:
> I'm trying to serve up a simple XMLRPC server as a windows service. I
> got it to run properly, I'm just not sure how to stop it properly.
> Most of the documentation/examples I found for this was from forums,
> so I'd love some links to relevant info also. Here's what I
> have...taken from the cookbook with the xmlrpc server added:
>
> def __init__(self, args):
> win32serviceutil.ServiceFramework.__init__(self, args)
> # Create an event which we will use to wait on.
> # The "service stop" request will set this event.
> self.hWaitStop = win32event.CreateEvent(None, 0, 0, None)
>
> def SvcStop(self):
> # Before we do anything, tell the SCM we are starting the stop
> process.
> self.ReportServiceStatus(win32service.SERVICE_STOP_PENDING)
>
> # quit the xmlrpc sever
> self.server.quit()
What is quit()? As the server may be processing a request I'd move any
finalization code below, after exiting the while loop.
>
> # And set my event.
> win32event.SetEvent(self.hWaitStop)
>
> def SvcDoRun(self):
> # Serve up the XMLRPC forever
> self.server =
> SimpleXMLRPCServer.SimpleXMLRPCServer(("10.0.1.6", 8000))
> self.server.register_instance(MyClass())
> self.server.serve_forever()
>
> win32event.WaitForSingleObject(self.hWaitStop)
The simplest solution is to replace serve_forever with a loop waiting on
hWaitStop:
while WaitForSingleObject(self.hWaitStop, 0)==WAIT_TIMEOUT:
self.server.handle_request()
Set the socket timeout to a reasonable value (you'll have to wait that
time before exiting). Also, a ThreadingTCPServer may be better if you
expect more than a request at a time. If you search past messages you may
find other ways.
--
Gabriel Genellina
--
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