时间:2021-07-01 10:21:17 帮助过:91人阅读
最近有个Python程序需要安装并作为Windows系统服务来运行,过程中碰到一些坑,整理了一下。
Python服务类
首先Python程序需要调用一些Windows系统API才能作为系统服务,具体内容如下:
#!/usr/bin/env python # -*- coding: utf-8 -*- import sys import time import win32api import win32event import win32service import win32serviceutil import servicemanager class MyService(win32serviceutil.ServiceFramework): _svc_name_ = "MyService" _svc_display_name_ = "My Service" _svc_description_ = "My Service" def init(self, args): self.log('init') win32serviceutil.ServiceFramework.init(self, args) self.stop_event = win32event.CreateEvent(None, 0, 0, None) def SvcDoRun(self): self.ReportServiceStatus(win32service.SERVICE_START_PENDING) try: self.ReportServiceStatus(win32service.SERVICE_RUNNING) self.log('start') self.start() self.log('wait') win32event.WaitForSingleObject(self.stop_event, win32event.INFINITE) self.log('done') except BaseException as e: self.log('Exception : %s' % e) self.SvcStop() def SvcStop(self): self.ReportServiceStatus(win32service.SERVICE_STOP_PENDING) self.log('stopping') self.stop() self.log('stopped') win32event.SetEvent(self.stop_event) self.ReportServiceStatus(win32service.SERVICE_STOPPED) def start(self): time.sleep(10000) def stop(self): pass def log(self, msg): servicemanager.LogInfoMsg(str(msg)) def sleep(self, minute): win32api.Sleep((minute*1000), True) if name == "main": if len(sys.argv) == 1: servicemanager.Initialize() servicemanager.PrepareToHostSingle(MyService) servicemanager.StartServiceCtrlDispatcher() else: win32serviceutil.HandleCommandLine(MyService)
pyinstaller打包
pyinstaller -F MyService.py
测试
# 安装服务 dist\MyService.exe install # 启动服务 sc start MyService # 停止服务 sc stop MyService # 删除服务 sc delete MyService
以上就是详解Python制作Windows系统服务的实例的详细内容,更多请关注Gxl网其它相关文章!