将Flask应用程序部署为Windows服务

Question

我正在使用此处找到的模板:是否可以在Windows中将Python脚本作为服务运行？如果可能,怎么样？

这是我的run.py,我按照上面链接中的说明安装了服务.

from app import app

import win32serviceutil
import win32service
import win32event
import servicemanager
import socket


class AppServerSvc (win32serviceutil.ServiceFramework):
    _svc_name_ = "Flask App"
    _svc_display_name_ = "Flask App"

    def __init__(self,args):
        win32serviceutil.ServiceFramework.__init__(self,args)
        self.hWaitStop = win32event.CreateEvent(None,0,0,None)
        socket.setdefaulttimeout(60)

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

    def SvcDoRun(self):
        servicemanager.LogMsg(servicemanager.EVENTLOG_INFORMATION_TYPE,
                              servicemanager.PYS_SERVICE_STARTED,
                              (self._svc_name_,''))
        self.main()

    def main(self):
        app.run(host = '192.168.1.6')

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

但是,当我尝试启动服务时,我收到消息:

"本地计算机上的Flask App服务已启动然后停止.如果某些服务未被其他服务或程序使用,则会自动停止."

知道我做错了什么吗？我尝试了各种用户帐户 - 我不认为这是一个权限问题.

谢谢!

Answer 1

我无法访问WSGIRequestHandler外面的Flask request,所以我使用Process.

import win32serviceutil
import win32service
import win32event
import servicemanager
from multiprocessing import Process

from app import app


class Service(win32serviceutil.ServiceFramework):
    _svc_name_ = "TestService"
    _svc_display_name_ = "Test Service"
    _svc_description_ = "Tests Python service framework by receiving and echoing messages over a named pipe"

    def __init__(self, *args):
        super().__init__(*args)

    def SvcStop(self):
        self.ReportServiceStatus(win32service.SERVICE_STOP_PENDING)
        self.process.terminate()
        self.ReportServiceStatus(win32service.SERVICE_STOPPED)

    def SvcDoRun(self):
        self.process = Process(target=self.main)
        self.process.start()
        self.process.run()

    def main(self):
        app.run()


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