Python win32服务
发布时间:2020-12-20 11:16:22 所属栏目:Python 来源:网络整理
导读:我有一个最小的 python win32服务service.py没有什么特别的: import win32serviceutilimport win32serviceimport win32eventclass SmallestPythonService(win32serviceutil.ServiceFramework): _svc_name_ = "SmallestPythonService" _svc_display_name_ =
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,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,},) 解决方法
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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