My Flask service stops because I logged out in Windows. When I reload my username and password, it works again. How can I make it always on, even when I reboot or log out?
import time
import random
import os
from pathlib import Path
from SMWinservice import SMWinservice
import sys
import win32serviceutil
import win32service
import win32event
import servicemanager
class FlaskService(SMWinservice):
_svc_name_ = "FlaskServiceSmartFactory01"
_svc_display_name_ = "Flask Service Smart Factory01"
_svc_description_ = "Python service framework Smart Factory01"
def start(self):
self.isrunning = True
def stop(self):
self.isrunning = False
def main(self):
os.chdir(os.environ['SMARTFACTORYPATH'])
os.system('C02_RestAPI.py')
if __name__ == '__main__':
if len(sys.argv) == 1:
servicemanager.Initialize()
servicemanager.PrepareToHostSingle(FlaskService)
servicemanager.StartServiceCtrlDispatcher()
else:
win32serviceutil.HandleCommandLine(FlaskService)
Windows services can run in background when Windows is running but you can't run it if Windows is off.
You can set this service to be load at Windows start if you want.
Related
I have created a test service file in pyhton:
import datetime
import win32serviceutil
import win32service
import win32event
import servicemanager
import socket
def writetofile():
DIR = "D:\prog\log.txt"
while True:
with open(DIR, 'a+') as file:
file.write(str(datetime.now()))
class AppServerSvc (win32serviceutil.ServiceFramework):
_svc_name_ = "TESTEST"
_svc_display_name_ = "TESTEST"
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):
servicemanager.LogMsg(servicemanager.EVENTLOG_INFORMATION_TYPE,
servicemanager.PYS_SERVICE_STARTED,
(self._svc_name_,''))
self.main()
def main(self):
writetofile()
if __name__ == '__main__':
win32serviceutil.HandleCommandLine(AppServerSvc)
I've also installed it but when I want to start in win services it stops every time with 1053 error.
I have tried to overwrite the register file (ServicesPipeTimeout) not worked. I set it up 180000 and nothing. My program ran for a while then stopped again.
I've installed everthing in cmd as admin (pip install pywin32), not worked.
And I've installed the service too in cmd (python main.py install) and I see it, but thats all. I cant start it, cant delete it cant do anything with it.
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
I'm trying to make windows service in python.
I have three source files: add_to_startup.py (adding my service to windows startup), run_as_a_service.py (runs my program as a service) and test.py (my program).
First i create .exe with pyinstaller:
pyinstaller -F --onefile --hidden-import=win32timezone run_as_a_service.py test.py add_to_startup.py
When i trying to start service error occurs:
Error 1053: The service did not respond in a timely fashion.
If i delete strings with test import and test.service() function, service works.
run_as_service.py:
import win32serviceutil
import win32service
import win32event
import servicemanager
import socket
import sys
import time
import test
class AppServerSvc (win32serviceutil.ServiceFramework):
_svc_name_ = "test"
_svc_display_name_ = "Test"
_svc_description_ = "test test"
def __init__(self,args):
win32serviceutil.ServiceFramework.__init__(self,args)
self.hWaitStop = win32event.CreateEvent(None,0,0,None)
socket.setdefaulttimeout(60)
def SvcStop(self):
self.isAlive = False
self.ReportServiceStatus(win32service.SERVICE_STOP_PENDING)
win32event.SetEvent(self.hWaitStop)
def SvcDoRun(self):
self.isAlive = True
servicemanager.LogMsg(servicemanager.EVENTLOG_INFORMATION_TYPE,
servicemanager.PYS_SERVICE_STARTED,
(self._svc_name_,''))
self.main()
def main(self):
while self.isAlive:
test.service()
time.sleep(5)
if __name__ == '__main__':
if len(sys.argv) == 1:
servicemanager.Initialize()
servicemanager.PrepareToHostSingle(AppServerSvc)
servicemanager.StartServiceCtrlDispatcher()
else:
win32serviceutil.HandleCommandLine(AppServerSvc)
test.py:
import add_to_startup
def service():
print("Hello world")
I am trying to write dbus server where I want to run some external shell program (grep here) to do the job.
when I do:
prompt$ server.py
then:
prompt$ client.py # works fine, ie. runs grep command in child process.
prompt$ client.py # ..., but second invocation produces following error message:
DBusException: org.freedesktop.DBus.Error.ServiceUnknown: The name org.example.ExampleService was not provided by any .service files
I am stuck. Are You able to help me?
here is server.py (client.py thereafter):
import gtk, glib
import os
import dbus
import dbus.service
import dbus.mainloop.glib
import subprocess
messages_queue=list()
grep_pid=0
def queue_msg(message):
global messages_queue
messages_queue.append(message)
return
def dequeue_msg():
global messages_queue,grep_pid
if grep_pid != 0:
try:
pid=os.waitpid(grep_pid,os.P_NOWAIT)
except:
return True
if pid[0] == 0:
return True
grep_pid=0
if len(messages_queue) == 0:
return True
else:
tekst=messages_queue.pop(0)
cmd="grep 'pp'"
print cmd
#works fine, when I do return here
#return True
grep_pid=os.fork()
if grep_pid != 0:
return True
os.setpgid(0,0)
pop=subprocess.Popen(cmd,shell=True,stdin=subprocess.PIPE)
pop.stdin.write(tekst)
pop.stdin.close()
pop.wait()
exit(0)
class DemoException(dbus.DBusException):
_dbus_error_name = 'org.example.Exception'
class MyServer(dbus.service.Object):
#dbus.service.method("org.example.ExampleInterface",
in_signature='', out_signature='')
def QueueMsg(self):
queue_msg("ppppp")
#dbus.service.method("org.example.ExampleInterface",
in_signature='', out_signature='')
def Exit(self):
mainloop.quit()
from dbus.mainloop.glib import threads_init
if __name__ == '__main__':
glib.threads_init()
threads_init()
dbus.mainloop.glib.DBusGMainLoop(set_as_default=True)
session_bus = dbus.SessionBus()
name = dbus.service.BusName("org.example.ExampleService", session_bus)
object = MyServer(session_bus, '/My')
glib.timeout_add_seconds(1, dequeue_msg)
mainloop = glib.MainLoop()
print "Running example service."
mainloop.run()
now client.py:
import sys
from traceback import print_exc
import dbus
def main():
bus = dbus.SessionBus()
try:
remote_object = bus.get_object("org.example.ExampleService",
"/My")
except dbus.DBusException:
print_exc()
sys.exit(1)
iface = dbus.Interface(remote_object, "org.example.ExampleInterface")
iface.QueueMsg()
if sys.argv[1:] == ['--exit-service']:
iface.Exit()
if __name__ == '__main__':
main()
You usually get this error message when you try to access a service that is no longer available. Check if your server is still running.
You can use d-feet to debug your dbus connections.
The error message about the missing .service file means that you need to create a service file in dbus-1/services.
For example:
# /usr/local/share/dbus-1/services/org.example.ExampleService.service
[D-BUS Service]
Name=org.example.ExampleService
Exec=/home/user1401567/service.py
A lot of tutorials don't include this detail (maybe .service files didn't use to be required?) But, at least on Ubuntu 12.04, dbus services can't be connected to without it.