如何保持Windows服务运行

Question

下面是我正在运行的Windows服务的框架.如果发生错误,将记录错误并可在事件查看器中查看.问题是,即使我已将恢复设置为在第一次,第二次和后续故障时重新启动服务,脚本也会退出并且不会再次重新启动.目前我几乎没有错误处理,因为我想看看事件查看器中可能出现的错误,以便我可以编写代码来相应地处理这些错误.

from win32api import CloseHandle, GetLastError, SetConsoleCtrlHandler
import os
import sys 
import time
import pythoncom
import win32serviceutil
import win32service
import win32event
import servicemanager
import socket

class AppServerSvc (win32serviceutil.ServiceFramework):
    _svc_name_ = "my_service_name"
    _svc_display_name_ = "my service"

    def __init__(self,args):
        win32serviceutil.ServiceFramework.__init__(self,args)
        SetConsoleCtrlHandler(lambda x: True, True)
        self.hWaitStop = win32event.CreateEvent(None,0,0,None)

    def SvcStop(self):
        self.ReportServiceStatus(win32service.SERVICE_STOP_PENDING)
        win32event.SetEvent(self.hWaitStop)
        self.run = False

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

    def main(self):
        while self.run == True
            pass          

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

编辑:

我试着尝试:除了自我.但结果仍然是相同的.当它崩溃时服务没有重新开始......请有任何想法吗？如果一个服务在崩溃的情况下无法重新启动,那么这个服务并不是那么有用......不妨将它作为.pyc运行它

编辑:

下面是我的脚本中可能出现的错误示例...我不相信这个错误消息特别有用,因为我试图实现的是重新启动服务但是这里的一个例子是一个例子.在没有重新启动的情况下崩溃我的服务的错误:

The instance's SvcRun() method failed 
Traceback (most recent call last):
  File "C:\Python27\lib\site-packages\win32\lib\win32serviceutil.py", line 806, in SvcRun
    self.SvcDoRun()
  File "C:\Some_Service.py", line 46, in SvcDoRun
    self.main()
  File "Some_Service.py", line 61, in main
    ser = self.open_serial_port()
  File "Some_Service.py", line 70, in open_serial_port
    serial_connection.open()
  File "C:\Python27\lib\site-packages\serial\serialwin32.py", line 56, in open
    raise SerialException("could not open port %s: %s" % (self.portstr, ctypes.WinError()))
SerialException: could not open port COM6: [Error 1225] The remote system refused the network connection. 
%2: %3

Answer 1

下面是一个服务,它除以零以引发错误.如果出现错误,将发送一个事件并使用该服务退出os._exit(-1).该值必须是0以外的任何值,以便Windows知道该服务没有很好地退出.

from win32api import CloseHandle, GetLastError, SetConsoleCtrlHandler
import os
import sys
import time

import win32serviceutil
import win32service
import win32event
import servicemanager

import traceback



class AppServerSvc (win32serviceutil.ServiceFramework):
    _svc_name_ = "test"
    _svc_display_name_ = "test"

    def __init__(self,args):
        win32serviceutil.ServiceFramework.__init__(self,args)
        SetConsoleCtrlHandler(lambda x: True, True)
        self.hWaitStop = win32event.CreateEvent(None,0,0,None)




    def SvcStop(self):
        self.ReportServiceStatus(win32service.SERVICE_STOP_PENDING)
        win32event.SetEvent(self.hWaitStop)
        self.run = False
    def SvcDoRun(self):

        servicemanager.LogMsg(servicemanager.EVENTLOG_INFORMATION_TYPE,
                              servicemanager.PYS_SERVICE_STARTED,
                              (self._svc_name_,''))
        self.run = True
        try: # try main
            self.main()
        except:
            servicemanager.LogErrorMsg(traceback.format_exc()) # if error print it to event log
            os._exit(-1)#return some value other than 0 to os so that service knows to restart


    def main(self):

        while self.run == True:
            time.sleep(30)         
            t = 1/0

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