An example Windows service implemented with pywin32 wrappers. #python #windows-service #pywin32

helloworld-win32-service.py

import win32serviceutil
import win32service
import win32event
import servicemanager
import socket
import time
import logging

logging.basicConfig(
    filename = 'c:\\Temp\\hello-service.log',
    level = logging.DEBUG, 
    format = '[helloworld-service] %(levelname)-7.7s %(message)s'
)

class HelloWorldSvc (win32serviceutil.ServiceFramework):
    _svc_name_ = "HelloWorld-Service"
    _svc_display_name_ = "HelloWorld Service"
    
    def __init__(self,args):
        win32serviceutil.ServiceFramework.__init__(self,args)
        self.stop_event = win32event.CreateEvent(None,0,0,None)
        socket.setdefaulttimeout(60)
        self.stop_requested = False

    def SvcStop(self):
        self.ReportServiceStatus(win32service.SERVICE_STOP_PENDING)
        win32event.SetEvent(self.stop_event)
        logging.info('Stopping service ...')
        self.stop_requested = True

    def SvcDoRun(self):
        servicemanager.LogMsg(
            servicemanager.EVENTLOG_INFORMATION_TYPE,
            servicemanager.PYS_SERVICE_STARTED,
            (self._svc_name_,'')
        )
        self.main()

    def main(self):
        logging.info(' ** Hello PyWin32 World ** ')
        # Simulate a main loop
        for i in range(0,50):
            if self.stop_requested:
                logging.info('A stop signal was received: Breaking main loop ...')
                break
            time.sleep(5)
            logging.info("Hello at %s" % time.ctime())
        return

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

Just a note:
If you are pretending to use the service with PyInstaller to generate a single file, modify the file according to this post.

if in the command prompt and were in the directory of this file,
and entering command
python helloworld-win32-service.py

this script will create the service just fine but when running i get a 1053 error "the service did not respond to the start or control request in a timely fashion"
i understand the thread you linked touches on it but could you update the example to help me figure it out? I have tried to run this script into a .EXE using pyinstaller, still no dice. thanks a bunch your code got me the farthest to understanding this process

I'm very curious as to what the purpose of the line 22 does in the __init__() method.

socket.setdefaulttimeout(60)

Does this service run every time the computer is booted? I mean does it autorun even before anyone logs in? Because I want to play music before signing in.

hmmm..
python .py install
Installing service <name_service>
Service installed
BUT not started, I have a message: "The service has not been started. (1062)"
what wrong ?

my service also not starting

my service also not starting

I've solved that problem by changing account that running the service.

One of the possible reasons might be inaccessibility of some DLLs.
For me it was pywintypes36.dll (and pythoncom36.dll).

faced the same issue solved it by finding out pywintypes36.dll pythoncom36.dll and placing it inside win32 with pythonservices.exe.

For future googlers finding this snippet, I had a heck of a time figuring out how to avoid the dreaded 1053 error when trying to start the service (from an executable compiled with pyinstaller).

This snippet appears to be missing the part that actually invokes the main() of the service class when the Windows Service framework runs the executable.

This tutorial is more complete:
https://www.codeproject.com/Articles/1115336/Using-Python-to-Make-a-Windows-Service

The difference is going from this:

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

to this

if __name__ == '__main__':
    if len(sys.argv) == 1:
        servicemanager.Initialize()
        servicemanager.PrepareToHostSingle(HelloWorldSvc)
        servicemanager.StartServiceCtrlDispatcher()
    else:
        win32serviceutil.HandleCommandLine(HelloWorldSvc)

This works because when the Service executes all it does is run the executable with no args. if len(sys.argv) == 1: checks how many command line args were supplied so it knows whether it's being run by the user with args such as install or otherwise being run as a service.

I hope this helps. Good luck!

I used command below to create the service successfullt.
sc create binpath= "C:\Python27\Python.exe " DisplayName= "" start= auto

The difference is going from this:

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

to this

if __name__ == '__main__':
    if len(sys.argv) == 1:
        servicemanager.Initialize()
        servicemanager.PrepareToHostSingle(HelloWorldSvc)
        servicemanager.StartServiceCtrlDispatcher()
    else:
        win32serviceutil.HandleCommandLine(HelloWorldSvc)

This works because when the Service executes all it does is run the executable with no args. if len(sys.argv) == 1: checks how many command line args were supplied so it knows whether it's being run by the user with args such as install or otherwise being run as a service.

I hope this helps. Good luck!

Thanks! You saved my day.

I created windows service on python and it's running properly.But when when I am trying stop or abort service in between , it giving error 1053. What can be the solution to stop the running service

For anyone else encountering a 1053 error with pyinstaller, it is also worth checking for other errors not necessarily related to the answer above.

In my case, turns out my program has relative paths for config files, etc.

Windows Services set C:\Windows\system32 as the working directory for the application.

This answer for Determining application path in a Python EXE generated by pyInstaller along with a os.chdir to change working directory worked for me:

# NOTE: for a --onefile executable, the path to the application is given by application_path = os.path.dirname(sys.executable) –
application_path = os.path.dirname(sys.executable)

# print(os.getcwd())  # will be C:\Windows\System32 when run as a service
# print('application_path ' + str(application_path))

# Change working directory to application path
os.chdir(application_path)

For anyone else encountering a 1053 error with pyinstaller, it is also worth checking for other errors not necessarily related to the answer above.

In my case, turns out my program has relative paths for config files, etc.

Windows Services set C:\Windows\system32 as the working directory for the application.

This answer for Determining application path in a Python EXE generated by pyInstaller along with a os.chdir to change working directory worked for me:

# NOTE: for a --onefile executable, the path to the application is given by application_path = os.path.dirname(sys.executable) –
application_path = os.path.dirname(sys.executable)

# print(os.getcwd())  # will be C:\Windows\System32 when run as a service
# print('application_path ' + str(application_path))

# Change working directory to application path
os.chdir(application_path)

How do you use it in the .py file?

I wrote alike program.
SERVICE_NAME: BBA Log mask
DISPLAY_NAME: BBA Log mask
TYPE : 10 WIN32_OWN_PROCESS
STATE : 2 START_PENDING
(NOT_STOPPABLE, NOT_PAUSABLE, IGNORES_SHUTDOWN)
WIN32_EXIT_CODE : 0 (0x0)
SERVICE_EXIT_CODE : 0 (0x0)
CHECKPOINT : 0x0
WAIT_HINT : 0x1388
It was able to start.
but it wansn't able to stop.
I didn't know the reason.

For anyone else encountering a 1053 error with pyinstaller, it is also worth checking for other errors not necessarily related to the answer above.

In my case, turns out my program has relative paths for config files, etc.

Windows Services set C:\Windows\system32 as the working directory for the application.

This answer for Determining application path in a Python EXE generated by pyInstaller along with a os.chdir to change working directory worked for me:

# NOTE: for a --onefile executable, the path to the application is given by application_path = os.path.dirname(sys.executable) –
application_path = os.path.dirname(sys.executable)

# print(os.getcwd())  # will be C:\Windows\System32 when run as a service
# print('application_path ' + str(application_path))

# Change working directory to application path
os.chdir(application_path)

why changig dir if you can os.path.join? that would be a little be faster for optimization