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，则会记录错误并可以在事件查看器中查看。问题是脚本只是退出并且不会再次重新启动，即使我已将恢复设置为在第一次、第二次和后续失败时重新启动服务。目前我几乎没有错误处理，因为我想查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：

我尝试尝试：除了 self.main 周围，但结果仍然相同。该服务在崩溃时没有重新启动......请有任何想法的人吗？如果在发生崩溃时无法重新启动服务，它就不是那么有用了……不妨将它作为 .pyc 运行

编辑：

下面是我的脚本中可能出现的错误示例...我不认为此错误消息特别有用，因为我试图实现的是让服务重新启动，但这里仍然是导致我的服务在没有重新启动的情况下崩溃的错误示例：

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

为什么不使用 try-except 呢？

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
            try :
                <your code>
            except :
                time.sleep(3)
                <may be an error log here>



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

Answer 2

以下是仅除以零以引发错误的服务。如果出现错误，则会发送一个事件并使用os._exit(-1) 退出服务。该值必须是除 0 以外的任何值，以便 windows 知道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)

Answer 3

我怀疑您的服务实际上并没有崩溃，而是在它有机会运行之前完成。您是否尝试过在main 函数中插入几秒钟的睡眠语句？