是否可以将Windows 10服务编写为python脚本而不涉及第三方软件,例如nssm或ActiveState python(不是说这两种方法都是错误的解决方案)。
仅
可能的?
答案 0 :(得分:1)
所以要使用this example我发现了一种最简单的方法来实现一个自给自足的python脚本而没有任何第三方编译器/服务:
安装win32模块包:
pip3 install pypiwin32
代码:
import win32serviceutil
import win32service
import win32event
import servicemanager
import socket
import sys
class TestService(win32serviceutil.ServiceFramework):
_svc_name_ = 'TestService'
_svc_display_name_ = 'TestService'
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):
f = open('D:\\test.txt', 'a')
rc = None
while rc != win32event.WAIT_OBJECT_0:
f.write('Test Service \n')
f.flush()
# block for 24*60*60 seconds and wait for a stop event
# it is used for a one-day loop
rc = win32event.WaitForSingleObject(self.hWaitStop, 24 * 60 * 60 * 1000)
f.write('shut down \n')
f.close()
if __name__ == '__main__':
if len(sys.argv) == 1:
servicemanager.Initialize()
servicemanager.PrepareToHostSingle(TestService)
servicemanager.StartServiceCtrlDispatcher()
else:
win32serviceutil.HandleCommandLine(TestService)