将python脚本作为服务运行时出现问题

ewo*_*wok 3 python windows service activepython

我已经使用这个问题来学习使用ActivePython将我的python脚本安装为服务.我可以安装我的脚本并将其删除,正如我所料,但我的问题是,一旦我开始运行脚本,我就失去了阻止它或删除它的能力.它只是永久运行,我无法摆脱它.链接问题的答案提到检查标志以停止脚本.有人可以解释如何做到这一点?

现在脚本做的很少.每隔几秒就会将行打印到一个文件中.我希望在继续处理更复杂的事情之前让它工作.

import pythoncom
import win32serviceutil
import win32service
import win32event
import servicemanager
import socket
import time


class AppServerSvc (win32serviceutil.ServiceFramework):
    _svc_name_ = "PythonTest"
    _svc_display_name_ = "Python Test"
    _svc_description_ = "This is a test of installing Python services"

    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):
        i=0
        while True:
            f=open("C:\\hello.txt","a")
            f.write("hello"+str(i)+"\n")
            f.close()
            i=i+1
            time.sleep(5)


if __name__ == '__main__':
    win32serviceutil.HandleCommandLine(AppServerSvc)
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fra*_*ca7 6

当然你无法阻止它,你陷入无限循环.更换:

time.sleep(5)
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有了这个:

if win32event.WaitForSingleObject(self.hWaitStop, 5000) == win32event.WAIT_OBJECT_0: 
    break
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应该做的伎俩.