最近在把Python的定时程序放进Windows服务时遇到了上述的问题‘错误1053：服务没有及时响应启动或控制请求’ 或者 ‘Error 1053: The service did not respond to the start or control request in a timely fashion’。
可以正常安装python PythonFileName.py install，
不能正常启动python PythonFileName.py start。
尝试使用python PythonFileName.py debug检查代码，没有发现代码问题，考虑应该是环境问题。
最终在StackOverflow的 Can't start Windows service written in Python (win32serviceutil) 中找到了 BuvinJ 的回答：“It might prove useful in this case to also add these directories to your system path: C:\Python27\Lib\site-packages\win32 and C:\Python27\Lib\site-packages\pywin32_system32. That will let you use pythonservice more easily. ”
解决办法
在cmd中输入
setx /M PATH "%PATH%;C:\Python27;C:\Users\Administrator\Anaconda2\Scripts;C:\Users\Administrator\Anaconda2\Lib\site-packages\pywin32_system32;C:\Users\Administrator\Anaconda2\Lib\site-packages\win32"
回车，增加两个路径的引用。然后就能启动服务了。
具体的Windows服务Python代码：
import sys
import win32serviceutil
import win32service
import win32event
import servicemanager
import winerror
class StockRecommendRunLoop(win32serviceutil.ServiceFramework):
_svc_name_ = "StockRecommendRunLoop"
_svc_display_name_ = "Stock Recommend Run Loop"
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):
self.ReportServiceStatus(win32service.SERVICE_RUNNING)
self.start_task()
win32event.WaitForSingleObject(self.hWaitStop, win32event.INFINITE)
if __name__ == '__main__':
if len(sys.argv) == 1:
try:
servicemanager.Initialize()
servicemanager.PrepareToHostSingle(StockRecommendRunLoop)
servicemanager.StartServiceCtrlDispatcher()
except win32service.error, details:
if details[0] == winerror.ERROR_FAILED_SERVICE_CONTROLLER_CONNECT:
win32serviceutil.usage()
else:
win32serviceutil.HandleCommandLine(StockRecommendRunLoop)
Reference:
Can't start Windows service written in Python (win32serviceutil)
Python Windows Service的方式运行
