Windows-service-python

[1] how to create a windows service in python (Python recipe)

Author: Alex Baker
Date: 7th July 2008
Description : Simple python program to generate wrap as a service based on example on the web, see link below.

http://essiene.blogspot.com/2005/04/python-windows-services.html

Usage : python aservice.py install
Usage : python aservice.py start
Usage : python aservice.py stop
Usage : python aservice.py remove

C:\>python aservice.py  --username <username> --password <PASSWORD> --startup auto install


import win32service
import win32serviceutil
import win32api
import win32con
import win32event
import win32evtlogutil
import os

class aservice(win32serviceutil.ServiceFramework):

   _svc_name_ = "aservice"
   _svc_display_name_ = "a service - it does nothing"
   _svc_description_ = "Tests Python service framework by receiving and echoing messages over a named pipe"

   def __init__(self, args):
       win32serviceutil.ServiceFramework.__init__(self, args)
       self.hWaitStop = win32event.CreateEvent(None, 0, 0, None)           

   def SvcStop(self):
       self.ReportServiceStatus(win32service.SERVICE_STOP_PENDING)
       win32event.SetEvent(self.hWaitStop)                    
        
   def SvcDoRun(self):
       import servicemanager      
       servicemanager.LogMsg(servicemanager.EVENTLOG_INFORMATION_TYPE,servicemanager.PYS_SERVICE_STARTED,(self._svc_name_, )) 
       self.timeout = 3000

   while 1:
       # Wait for service stop signal, if I timeout, loop again
       rc = win32event.WaitForSingleObject(self.hWaitStop, self.timeout)
       # Check to see if self.hWaitStop happened
       if rc == win32event.WAIT_OBJECT_0:
           # Stop signal encountered
           servicemanager.LogInfoMsg("aservice - STOPPED")
           break
       else:
           servicemanager.LogInfoMsg("aservice - is alive and well")   


def ctrlHandler(ctrlType):
    return True
                 
if __name__ == '__main__':   
    win32api.SetConsoleCtrlHandler(ctrlHandler, True)   
    win32serviceutil.HandleCommandLine(aservice)