Windows XMLRPC Service

[half.i...@gmail.com]

Hi,

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:

import win32serviceutil
import win32service
import win32event

import SimpleXMLRPCServer

class MyClass(object):
def hello(self):
return "Hello World!"

class SmallestPythonService(win32serviceutil.ServiceFramework):
_svc_name_ = "PythonXMLRPC"
_svc_display_name_ = "PythonXMLRPC"

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()

# 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)

if __name__=='__main__':
win32serviceutil.HandleCommandLine(SmallestPythonService)

~Sean

[Gabriel Genellina]

En Mon, 18 Jun 2007 00:25:25 -0300, <half.i...@gmail.com> 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

[half.i...@gmail.com]

On Jun 18, 2:16 am, "Gabriel Genellina" <gagsl-...@yahoo.com.ar>
wrote:

> Gabriel Genellina- Hide quoted text -
>
> - Show quoted text -

I can't quite figure out where to set the "socket timeout". I tried
setting win32event.WAIT_TIMEOUT, but I'm pretty sure that's not the
variable you were talking about. I did manage to make it multi-
threaded by incorporating a different recipe, and I'm beginning to
understand the control flow a bit better, but it doesn't seem to be
doing what I expect. When SvcStop() is executed and calls
win32event.SetEvent(self.hWaitStop), the while loop should break as
win32event.WaitForSingleObject(self.hWaitStop, 0) returns zero at this
point. But it doesn't do that. What am I missing?

import win32serviceutil
import win32service
import win32event

import SocketServer
from SimpleXMLRPCServer import
SimpleXMLRPCServer,SimpleXMLRPCRequestHandler

# Threaded mix-in
class
AsyncXMLRPCServer(SocketServer.ThreadingMixIn,SimpleXMLRPCServer):
pass

class MyClass(object):
def hello(self):
return "Hello World"

class SmallestPythonService(win32serviceutil.ServiceFramework):

_svc_name_ = "PythonXMLRPC"
_svc_display_name_ = "PythonXMLRPC"

def __init__(self, args):

win32serviceutil.ServiceFramework.__init__(self, args)
# Create an event which we will use to wait on.

self.hWaitStop = win32event.CreateEvent(None, 0, 0, None)

import socket
localhost = socket.gethostbyname(socket.gethostname())
self.server = AsyncXMLRPCServer((localhost, 8000),
SimpleXMLRPCRequestHandler)

def SvcStop(self):
self.ReportServiceStatus(win32service.SERVICE_STOP_PENDING)
win32event.SetEvent(self.hWaitStop)
#print "EVENT:",
win32event.WaitForSingleObject(self.hWaitStop, 0) # returns 0 here

def SvcDoRun(self):
self.server.register_instance(MyClass())

#win32event.WAIT_TIMEOUT = 2 --- This just makes the loop
never execute because
# the WaitFor... part always returns 258

while win32event.WaitForSingleObject(self.hWaitStop, 0) ==
win32event.WAIT_TIMEOUT:
self.server.handle_request()

if __name__ == '__main__':
win32serviceutil.HandleCommandLine(SmallestPythonService)

Thanks for any help!

~Sean

[Gabriel Genellina]

En Tue, 19 Jun 2007 03:45:19 -0300, <half.i...@gmail.com> escribió:

> I can't quite figure out where to set the "socket timeout". I tried
> setting win32event.WAIT_TIMEOUT, but I'm pretty sure that's not the
> variable you were talking about. I did manage to make it multi-
> threaded by incorporating a different recipe, and I'm beginning to
> understand the control flow a bit better, but it doesn't seem to be
> doing what I expect. When SvcStop() is executed and calls
> win32event.SetEvent(self.hWaitStop), the while loop should break as
> win32event.WaitForSingleObject(self.hWaitStop, 0) returns zero at this
> point. But it doesn't do that. What am I missing?

May be because you didn't set correctly the socket timeout. See the
comments below.

>
> def SvcStop(self):
> self.ReportServiceStatus(win32service.SERVICE_STOP_PENDING)
> win32event.SetEvent(self.hWaitStop)
> #print "EVENT:",
> win32event.WaitForSingleObject(self.hWaitStop, 0) # returns 0 here

That's OK, since you have set the event.

> def SvcDoRun(self):
> self.server.register_instance(MyClass())
>
> #win32event.WAIT_TIMEOUT = 2 --- This just makes the loop
> never execute because
> # the WaitFor... part always returns 258

WAIT_TIMEOUT is 258. How do you see it is 2?
For example, see <http://msdn2.microsoft.com/en-us/library/ms681382.aspx>.
Python 2.5.1 + pywin32 210 prints this on my PC:
py> import win32event
py> win32event.WAIT_TIMEOUT
258

> while win32event.WaitForSingleObject(self.hWaitStop, 0) ==
> win32event.WAIT_TIMEOUT:
> self.server.handle_request()

The loop above should keep running until hWaitStop is set, with a maximum
wait time (inside handle_request) corresponding to the socket timeout
value.
You can either:
- use socket.setdefaulttimeout() (in __init__, by example) before anything
else. This will set a global timeout for all sockets.
- modify the socket instance. Just add this method to your AsyncServer:
def server_activate(self):
SimpleXMLRPCServer.server_activate(self)
self.socket.settimeout(15) # for 15 secs

--
Gabriel Genellina

[half.i...@gmail.com]

On Jun 19, 10:21 am, "Gabriel Genellina" <gagsl-...@yahoo.com.ar>
wrote:

I meant here that *if* I set the WAIT_TIMEOUT to 2, then I see that
behavior.

> - use socket.setdefaulttimeout() (in __init__, by example) before anything
> else. This will set a global timeout for all sockets.

That was the one that did it.

Thank you again Gabriel. I'll post back with something complete.

~Sean

[Gabriel Genellina]

En Tue, 19 Jun 2007 14:57:10 -0300, <half.i...@gmail.com> escribió:

>> > #win32event.WAIT_TIMEOUT = 2 --- This just makes the loop
>> > never execute because
>> > # the WaitFor... part always returns 258
>>
>> WAIT_TIMEOUT is 258. How do you see it is 2?

>> py> import win32event
>> py> win32event.WAIT_TIMEOUT
>> 258
>
> I meant here that *if* I set the WAIT_TIMEOUT to 2, then I see that
> behavior.

Ah, ok! I should have stated clearly that WAIT_TIMEOUT is a Windows
predefined constant, not your desired timeout value.

> Thank you again Gabriel. I'll post back with something complete.

Yes, please, a working example would be nice for future readers...

--
Gabriel Genellina

[half.i...@gmail.com]

On Jun 19, 12:32 pm, "Gabriel Genellina" <gagsl-...@yahoo.com.ar>
wrote:

For posterity...

import sys

import win32serviceutil
import win32service
import win32event

import win32evtlogutil
import servicemanager

import SocketServer, socket
from SimpleXMLRPCServer import
SimpleXMLRPCServer,SimpleXMLRPCRequestHandler

# Threaded mix-in
class
AsyncXMLRPCServer(SocketServer.ThreadingMixIn,SimpleXMLRPCServer):
pass

import XMLRPC_funcs # module containing the functions wrapped in a
class

class XMLRPCservice(win32serviceutil.ServiceFramework):

_svc_name_ = "PythonXMLRPC"
_svc_display_name_ = "PythonXMLRPC"

_svc_description_ = "Multi-threaded Python XMLRPC Server"

def __init__(self, args):
# set the timeout so the service can stop...Otherwise it hangs
forever
socket.setdefaulttimeout(15)

win32evtlogutil.AddSourceToRegistry(self._svc_display_name_,
sys.executable, "Application")
win32serviceutil.ServiceFramework.__init__(self, args)

# Create an event which we will use to wait on.
self.hWaitStop = win32event.CreateEvent(None, 0, 0, None)

localhost = socket.gethostbyname(socket.gethostname())
self.server = AsyncXMLRPCServer((localhost, 8000),
SimpleXMLRPCRequestHandler)

def SvcStop(self):
self.ReportServiceStatus(win32service.SERVICE_STOP_PENDING)
#send the stop event
win32event.SetEvent(self.hWaitStop)

def SvcDoRun(self):
# log a start msg

servicemanager.LogMsg(servicemanager.EVENTLOG_INFORMATION_TYPE,
servicemanager.PYS_SERVICE_STARTED,
(self._svc_name_, ' (%s)' %
self._svc_name_))

self.server.register_instance(XMLRPC_funcs.XMLRPC_funcs())

# handle requests until the stop event is received

while win32event.WaitForSingleObject(self.hWaitStop, 0) ==
win32event.WAIT_TIMEOUT:
self.server.handle_request()

# log a stopped msg
win32evtlogutil.ReportEvent(self._svc_name_,
servicemanager.PYS_SERVICE_STOPPED,
0,

servicemanager.EVENTLOG_INFORMATION_TYPE,
(self._svc_name_,""))

if __name__ == '__main__':
win32serviceutil.HandleCommandLine(XMLRPCservice)