I am writing a python win32 service below are snippet of my code when i compile the service it works but i need to go to services.msc and start it manually.
Is there an option when i install the serivce by : myservice.exe install it will starts automaticly ?
below are snippet of my code :
import win32serviceutil
import win32service
import win32event
class SmallestPythonService(win32serviceutil.ServiceFramework):
_svc_name_ = "ser_name"
_svc_display_name_ = "ser_descryption"
#_svc_description_='ddd'
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):
win32event.WaitForSingleObject(self.hWaitStop, win32event.INFINITE)
if __name__=='__main__':
win32serviceutil.HandleCommandLine(SmallestPythonService)
Use myservice.exe --startup=auto install to install the service and set it to be started automatically.
I would take a look at this ActiveState recipe. It's a wrapper around the win32serviceutil that shows how to auto-start the service.
You can use sc.exe with the create command.
sc create MyPyService binPath= "C:\myservice.exe" DisplayName= "Some Python Service"
More on this at Microsoft KB251192.
win32serviceutil also has a InstallService() function you might be able to use.
#Maciejg doesn't work for me, here the solution to start automaticaly my service builded with py2exe :
myservice.exe -auto -install
Related
I know this question was asked so many times. I read all those questions but i didn't find out my problem's solution. My issue is that i have created below window service with help of this link. How do you run a Python script as a service in Windows?
and I am running this service from command prompt.
Here is my python script that i need to run as a service.
import traceback
import win32serviceutil
import win32service
import win32event
import servicemanager
import socket
import getpass
import json
import pathlib
import urllib.request
import sys
from time import time , sleep
import uuid
from urllib.request import urlopen , Request
from urllib.parse import urlencode , quote_plus
import subprocess
import threading
import requests
from requests.sessions import session
import os
import cgitb
import logging
class PythonService(win32serviceutil.ServiceFramework):
_svc_name_ = "PCSIGN"
_svc_display_name_ = "PC SIGN"
def __init__(self, args):
win32serviceutil.ServiceFramework.__init__(self, args)
self.hWaitStop = win32event.CreateEvent(None,0, 0, None)
socket.setdefaulttimeout(60)
def SvcStop( self ):
self.ReportServiceStatus(win32service.SERVICE_STOP_PENDING)
win32event.SetEvent(self.hWaitStop)
def SvcDoRun( self ):
win32event.WaitForSingleObject(self.hWaitStop, win32event.INFINITE)
def main( ):
while True:
file = open ( 'geek.txt' , 'a+' )
file.write("hello world")
file.close()
if __name__ == '__main__':
win32serviceutil.HandleCommandLine(PythonService)
PythonService.main()
As you can see in the main function there is an infinite loop in which I am opening a file and writing hello world in it.
I install & start my service from command prompt.
C:\wamp64\www\project\python\New folder>start /min python testing.py install
C:\wamp64\www\project\python\New folder>start /min python testing.py start
after that service installed and start working properly and another window appear.
It also makes an entry successfully in Services.
But the issue here is when i close the above window of python console it stop writing hello world in the file don't know why kindly how can i make it persistent so that whenever system restarted my script start working automatically and start writing hello world in the file
Just a few days back I have successfully sorted out my issue I just made some changes to my code and it started working properly. I am only posting these answers for someone like me who got this issue in the future he can easily sort it out.
This is the new code:
def WriteToFile():
while True:
file = open ( "C:\\file.txt" , "w" )
now = datetime.now()
now = now.strftime("%B %d, %y %H:%M:%S")
file.write(now)
class Pythonservice(win32serviceutil.ServiceFramework):
_svc_name_ = 'PC-Service'
_svc_display_name_ = 'PC-Service'
_svc_description_ = 'Freindly Service'
#classmethod
def parse_command_line(cls):
win32serviceutil.HandleCommandLine(cls)
def __init__(self, args):
win32serviceutil.ServiceFramework.__init__(self, args)
self.hWaitStop = win32event.CreateEvent(None, 0, 0, None)
socket.setdefaulttimeout(60)
def SvcStop(self):
self.stop()
self.ReportServiceStatus(win32service.SERVICE_STOP_PENDING)
win32event.SetEvent(self.hWaitStop)
def SvcDoRun(self):
self.start()
servicemanager.LogMsg(servicemanager.EVENTLOG_INFORMATION_TYPE,
servicemanager.PYS_SERVICE_STARTED,
(self._svc_name_, ''))
self.main()
def start(self):
self.isrunning = True
def stop(self):
self.isrunning = False
def main(self):
WriteToFile()
if __name__ == '__main__':
Pythonservice.parse_command_line()
After these changes, I opened up a command prompt as administrator and type this command to install my service.
C:\wamp64\www\project\python\New folder>python testing.py install
I got the success message. After successful installation, I started my service using this command
C:\wamp64\www\project\python\New folder>python testing.py start
and service started successfully I confirmed from service manager as well whether my service is running or not and it was running after restarting my pc service was still in running state.
If I try to do this it doesn't generate the warning:
Test.py install
Test.py start
Otherwise if I use pyinstaller:
pyinstaller -F --hidden-import=win32timezone Test.py
And then I try to do:
Test.exe install
Test.exe start
I see this warning in the event log:
A service process other than the one started by Service Control Manager connected when the TestService service started. Service Control Manager started process 5328 and connected process 1512.
Note that if the service is configured to start inside a debugger, this behavior is expected.
Script:
import servicemanager
import socket
import sys
import win32event
import win32service
import win32serviceutil
class TestService(win32serviceutil.ServiceFramework):
_svc_name_ = "TestService"
_svc_display_name_ = "Test Service"
_svc_description_ = "My service description"
def __init__(self, args):
win32serviceutil.ServiceFramework.__init__(self, args)
self.hWaitStop = win32event.CreateEvent(None, 0, 0, None)
socket.setdefaulttimeout(60)
def SvcStop(self):
self.ReportServiceStatus(win32service.SERVICE_STOP_PENDING)
win32event.SetEvent(self.hWaitStop)
def SvcDoRun(self):
rc = None
while rc != win32event.WAIT_OBJECT_0:
with open('C:\\TestService.log', 'a') as f:
f.write('test service running...\n')
rc = win32event.WaitForSingleObject(self.hWaitStop, 5000)
if __name__ == '__main__':
if len(sys.argv) == 1:
servicemanager.Initialize()
servicemanager.PrepareToHostSingle(TestService)
servicemanager.StartServiceCtrlDispatcher()
else:
win32serviceutil.HandleCommandLine(TestService)
How can I start the service without this warning appearing?
This is only a warning of windows complaining that the service fired two process.
This is the normal behaviour of pyinstaller executables. You should ignore this kind of warning.
See the docs
Using Python 3.7, Windows 10 Pro, Pywin32
I have a test script that starts a service and pushes some basic lines into a log file as the different commands are issued. Code is as follows:
import win32serviceutil
import win32service
import win32event
import servicemanager
import socket
import logging
class AppServerSvc(win32serviceutil.ServiceFramework):
_svc_name_ = "TestService"
_svc_display_name_ = "Test Service"
_svc_description_ = "New Test Service"
logging.basicConfig(filename='search_server.log', level=logging.INFO)
logging.info('Class opened')
def __init__(self, args):
logging.basicConfig(filename='search_server.log', level=logging.INFO)
logging.info('Init')
win32serviceutil.ServiceFramework.__init__(self, args)
self.hWaitStop = win32event.CreateEvent(None, 0, 0, None)
socket.setdefaulttimeout(60)
def SvcStop(self):
logging.basicConfig(filename='search_server.log', level=logging.INFO)
logging.info('Stop')
self.ReportServiceStatus(win32service.SERVICE_STOP_PENDING)
win32event.SetEvent(self.hWaitStop)
def SvcDoRun(self):
logging.basicConfig(filename='search_server.log', level=logging.INFO)
logging.info('Run')
servicemanager.LogMsg(servicemanager.EVENTLOG_INFORMATION_TYPE,
servicemanager.PYS_SERVICE_STARTED,
(self._svc_name_, ''))
self.main()
def main(self):
print("running")
logging.basicConfig(filename='search_server.log', level=logging.INFO)
logging.info('Main')
if __name__ == '__main__':
logging.basicConfig(filename='search_server.log', level=logging.INFO)
logging.info('Calling Handle Command Line')
win32serviceutil.HandleCommandLine(AppServerSvc)
I have gone through the basic trouble shooting with this, and the service is installing, starting, restarting and being removed without any errors. However I am expecting the log file to receive basic output to show the functions are being hit, and it isn't.
The calls I am making at the admin command prompt:
C:\PythonScripts\SearchServer>python servicetest.py install
Installing service TestService
Service installed
C:\PythonScripts\SearchServer>python servicetest.py start
Starting service TestService
C:\PythonScripts\SearchServer>python servicetest.py restart
Restarting service TestService
C:\PythonScripts\SearchServer>python servicetest.py remove
Removing service TestService
Service removed
C:\PythonScripts\SearchServer>
Contents of log file:
INFO:root:Class opened
INFO:root:Calling Handle Command Line
INFO:root:Class opened
INFO:root:Calling Handle Command Line
INFO:root:Class opened
INFO:root:Calling Handle Command Line
INFO:root:Class opened
INFO:root:Calling Handle Command Line
As you can see the service is being hit every time a command is issued, however I'd expect the internal functions to be called too. Being new to both services and Python I'm wondering if I've missed anything? I'm assuming the function names are predefined and don't need me to set up delegation to access them. Its not something I've seen in any of the questions I've come across.
I am of course assuming that these functions are supposed to be hit and they are being hit and are capable of creating logs?
Any help gratefully received.
There are a few problems with the code:
logging.basicConfig() should be called only once, if called again it won't have any effect.
Class definition will be called first in your code, even before block if __name__ == '__main__': because of natural flow of code. Due to this, whatever you set in logging.basicConfig() in class definition will become final for whole script. It is not an ideal place for this setting, so should be moved elsewhere (at the top, outside the class preferably).
filename parameter passed in the logging.basicConfig should be the absolute file path, because once service starts running, its current path won't be the same as the script, so logging won't be able to find out the log file. (current working directory for the service would become something like C:\Python37\lib\site-packages\win32).
(Optional): Try not to use root logging config at all, it's better to have an instance of logger for your own use.
After all these changes, the script will look like this:
import win32serviceutil
import win32service
import win32event
import servicemanager
import socket
import logging.handlers
log_file_path = "" # mention full path here
mylogger = logging.getLogger("TestLogger")
mylogger.setLevel(logging.INFO)
handler = logging.handlers.RotatingFileHandler(log_file_path)
mylogger.addHandler(handler)
class AppServerSvc(win32serviceutil.ServiceFramework):
_svc_name_ = "TestService"
_svc_display_name_ = "Test Service"
_svc_description_ = "New Test Service"
mylogger.info('Class opened')
def __init__(self, args):
mylogger.info('Init')
win32serviceutil.ServiceFramework.__init__(self, args)
self.hWaitStop = win32event.CreateEvent(None, 0, 0, None)
socket.setdefaulttimeout(60)
def SvcStop(self):
mylogger.info('Stop')
self.ReportServiceStatus(win32service.SERVICE_STOP_PENDING)
win32event.SetEvent(self.hWaitStop)
def SvcDoRun(self):
mylogger.info('Run')
servicemanager.LogMsg(servicemanager.EVENTLOG_INFORMATION_TYPE,
servicemanager.PYS_SERVICE_STARTED,
(self._svc_name_, ''))
self.main()
def main(self):
print("running")
mylogger.info('Main')
if __name__ == '__main__':
mylogger.info('Calling Handle Command Line')
win32serviceutil.HandleCommandLine(AppServerSvc)
Output:
Class opened
Init
Run
Main
Class opened
Calling Handle Command Line
I've had unexplainable problems with python logging, I solved them by setting up the logging right at the beginning of the program:
import logging
logging.basicConfig(filename='convert_st.log', level=logging.INFO)
logging.info('Started')
import all_other_packages
import...
...
def main:
# main comes here
...
if __name__ == '__main__':
main()
I'm getting an error when using a python exe generated from pyinstaller, to create a windows service. The error message may be innocuous, and doesn't seem to effect the operation of the service, but I'm not sure if there are other problems going on behind the scenes. I'm using the pywin32 libraries to install the application as a windows service. I should note that I don't get this error when installing from the python script itself, using PythonService.exe from pywin32, only from the executable generated with pyinstaller.
When using pyinstaller, I can generate the exe from my windows service code and install it, no problem. I can also start the service, no problem. I can even stop the service and the application appears to shut down properly. However, once I've initiated the stop, I get the following error on the console while running the win32traceutil.py:
"Exception KeyError: KeyError(2244,) in <module 'threading' from '2\build\pyi.win32\agentservice\outPYZ1.pyz/threading'> ignored"
No errors are recorded to the event log. I've been able to trace it back to the python logging module. Simply importing the logging module seems to cause my problem. Commenting out the import eliminates the error. It seems pretty clear to me that this is causing the problem, but I find it strange that pyinstaller would have issues with a module in the standard library. Has anyone else run into this?
I'm running Python 2.6.6, Pyinstaller 1.5.1, Build 217 of Pywin32. I'm on Windows XP.
And a stripped down version of my code:
import win32service
import win32serviceutil
import win32event
import win32evtlogutil
import win32traceutil
import servicemanager
import sys
import os
import time
class myservice(win32serviceutil.ServiceFramework):
_svc_name_ = "5"
_svc_display_name_ = "5"
_svc_deps_ = ["EventLog"]
def __init__(self, args):
self.isAlive = True
win32serviceutil.ServiceFramework.__init__(self, args)
self.hWaitStop = win32event.CreateEvent(None, 0, 0, None)
def event_log(self, msg):
servicemanager.LogInfoMsg(str(msg))
def SvcStop(self):
# tell Service Manager we are trying to stop (required)
self.ReportServiceStatus(win32service.SERVICE_STOP_PENDING)
print "svcstop: stopping service, setting event"
# set the event to call
win32event.SetEvent(self.hWaitStop)
print "svcstop: ending svcstop"
def SvcDoRun(self):
print "we are starting the service..."
self.event_log("Starting %s" % self._svc_name_)
############ IF LOGGING IS COMMENTED OUT, THE ERROR GOES AWAY################
import logging
print "svcdorun: waiting for object"
win32event.WaitForSingleObject(self.hWaitStop,win32event.INFINITE)
print "svcdorun: return from function"
if __name__ == '__main__':
if len(sys.argv)==1:
import win32traceutil
print "service is starting..."
#servicemanager.Initialize()
servicemanager.Initialize('backup service', None)
servicemanager.PrepareToHostSingle(myservice)
# Now ask the service manager to fire things up for us...
servicemanager.StartServiceCtrlDispatcher()
print "service done!"
else:
win32serviceutil.HandleCommandLine(myservice)
I have a minimal python win32 service service.py that does nothing special:
import win32serviceutil
import win32service
import win32event
class SmallestPythonService(win32serviceutil.ServiceFramework):
_svc_name_ = "SmallestPythonService"
_svc_display_name_ = "display service"
# _svc_description_='ddd'
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):
win32event.WaitForSingleObject(self.hWaitStop, win32event.INFINITE)
if __name__=='__main__':
win32serviceutil.HandleCommandLine(SmallestPythonService)
When I run:
service.py install
service.py start
it works fine but when I compile the service.py file with py2exe to service.exe and run the following:
service.exe install
service.exe start [or trying to restart the service from the Services.msc]
I get this message:
Could not start the service name service on Local Computer.
Error 1053: The service did not respond to the start or control request in a timely fashion
How can I resolve this problem?
Also here the distutil code:
from distutils.core import setup
import py2exe
py2exe_options = {"includes": ['decimal'],'bundle_files': 1}
setup(console=[{"script":'Service.py'}],
options={"py2exe": py2exe_options},
zipfile = None,
},
)
Replace your: setup(console=[{"script":'Service.py'}] with setup(service=[{"script":'Service.py'}]. Instead of console use service.
try this setup:
py2exe_options = {"includes": ['decimal'],'bundle_files': 1}
setup(
service=[{'modules':'Service.py','cmdline_style':'pywin32','description':'your service description'}],
options={'py2exe':py2exe_options},
zipfile=None)
A quick google came up with this: http://islascruz.org/html/index.php?gadget=StaticPage&action=Page&id=6
It has Italian comments, but I can help you translate some stuff if you don't know Italian.
To truly debug your problem, I guess we will need to see your setup.py distutils script...
You probably might be missing the right PATH for finding all the DLLs required by the service. Usually the service gets installed as a 'LocalSystem' service so you need to add the PATH to the System (and not to User).
Try adding c:\python27 (or whatever the path to your python dlls is) to the SYSTEM PATH, restart the computer and check if it now starts fine.