Python win32服务

2019年8月4日 30次阅读

我有一个最小的
python win32服务service.py没有什么特别的：

import win32serviceutil
import win32service
import win32event

class SmallestPythonService(win32serviceutil.ServiceFramework):
    _svc_name_ = "SmallestPythonService"
    _svc_display_name_ = "display service"
    # _svc_description_='ddd'

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

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

    def SvcDoRun(self):     
         win32event.WaitForSingleObject(self.hWaitStop, win32event.INFINITE)

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

当我跑：

 service.py install
 service.py start

它工作正常,但当我使用py2exe将service.py文件编译为service.exe并运行以下代码时：

service.exe install
service.exe start [or trying to restart the service  from the Services.msc]

我收到这条消息：

Could not start the  service name service on Local Computer.
Error 1053: The service did not respond to the start or control request in a timely fashion

我该如何解决这个问题？

此处还有distutil代码：

from distutils.core import setup
import py2exe

py2exe_options = {"includes": ['decimal'],'bundle_files': 1}

setup(console=[{"script":'Service.py'}], 
    options={"py2exe": py2exe_options}, 
    zipfile = None,
    },
 )

最佳答案 用setup(service = [{“script”：’Service.py’}]替换你的：setup(console = [{“script”：’Service.py’}].而不是控制台使用服务.

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