如何使Windows服务保持运行

Question

以下是我正在运行的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。问题是脚本退出并且不会重新启动，即使我已将恢复设置为在第一次，第二次和随后的故障时重新启动服务。目前我的错误处理很少，因为我想查看事件查看器中可能出现的错误，以便我可以编写相应的代码来处理这些错误。

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：

我想尝试：除了围绕self.main但是结果还是一样。该服务在崩溃时不会重新启动...请任何人有任何想法？如果它不能在发生碰撞的情况下，重新启动服务是不是所有的有用......还不如运行它作为一个.pyc文件

编辑：

下面

是中可能出现的错误的例子我的脚本......我不相信这个错误消息是特别有用的，因为我试图实现的是让服务重新启动，但是这里没有一个是在没有重新启动的情况下崩溃我的服务的错误的示例：

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

下面是一个以零分出错的服务。如果发生错误，则会发送事件并使用os._exit(-1)退出服务。该值必须是0以外的任何值，以便窗口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 2

我怀疑你的服务实际上并没有崩溃，而是在它有机会运行之前完成。您是否尝试在main函数中插入几秒钟的睡眠声明？

Answer 3

为什么不使用try-except？

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)