Python Win32服务除了在调试模式下启动

Question

我有一个Django应用程序，我正在尝试通过Windows Server上的Cherrypy运行。我在服务创建脚本中的代码以下。

import os
import sys

service_directory = os.path.dirname(__file__)
source_directory = os.path.abspath(service_directory)
os.chdir(source_directory)
venv_base = os.path.abspath(os.path.join(source_directory, ".venv"))
print(venv_base)
sys.path.append(".")
old_os_path = os.environ['PATH']
os.environ['PATH'] = os.path.join(venv_base, "Scripts")+ os.pathsep + old_os_path
site_packages = os.path.join(venv_base, "Lib", "site-packages")
prev_sys_path = list(sys.path)
import site
site.addsitedir(site_packages)
sys.real_prefix = sys.prefix
sys.prefix = venv_base
new_sys_path = list()
for item in list(sys.path):
    if item not in prev_sys_path:
        new_sys_path.append(item)
        sys.path.remove(item)
sys.path[:0] = new_sys_path

import pathlib
import subprocess
import traceback
import servicemanager
import win32event
import win32service
import win32serviceutil
from win32api import SetConsoleCtrlHandler

class AppServerSvc(win32serviceutil.ServiceFramework):
    """
    Service Create class
    """
    _svc_name_ = "MyService"  # service name
    _svc_display_name_ = "MyService"  # display name

    def __init__(self, args):
        """
        Constructor
        """
        win32serviceutil.ServiceFramework.__init__(self, args)
        SetConsoleCtrlHandler(lambda x: True, True)
        self.hWaitStop = win32event.CreateEvent(None, 0, 0, None)

    def SvcStop(self):
        """
        Function to stop the windows service
        """
        self.ReportServiceStatus(win32service.SERVICE_STOP_PENDING)
        win32event.SetEvent(self.hWaitStop)
        self.proc.terminate()

    def SvcDoRun(self):
        """
        Function to create and start the windows service
        """
        self.ReportServiceStatus(win32service.SERVICE_START_PENDING)
        servicemanager.LogMsg(servicemanager.EVENTLOG_INFORMATION_TYPE,
                              servicemanager.PYS_SERVICE_STARTED,
                              (self._svc_name_, ''))
        try:
            self.main()

        except Exception as exp_msg:
            servicemanager.LogErrorMsg(traceback.format_exc())
            os._exit(-1)

    def main(self):
        """
        Script to run as the windows service
        """
        #env_file.load(f'{os.path.join(pathlib.Path(__file__).parent.absolute(), "config.env")}')
        #settings.set_env_variables()
        self.ReportServiceStatus(win32service.SERVICE_RUNNING)
        self.proc = subprocess.Popen(f"python"
                                     f" {os.path.join(pathlib.Path(__file__).parent.absolute(), 'cherryd.py')} 1"
                                     , stdout=subprocess.PIPE, stderr=subprocess.STDOUT,
                                     stdin=subprocess.DEVNULL)
        stdout, stderr = self.proc.communicate()
        print(stdout)
        if (self.proc.poll() == 0):
            servicemanager.LogMsg(f"Successfully started the process with PID: {self.proc.pid}")
        else:
            servicemanager.LogErrorMsg(f"Failed to start the services")


if __name__ == '__main__':
    # Start of the process
    win32serviceutil.HandleCommandLine(AppServerSvc)

我可以使用python service.py install安装此服务，这里没有问题。 当我启动服务python service.py start我获得错误错误启动服务：服务没有及时响应开始或控制请求。

当我执行python service.py debug，一切正常，我能够使用FQDN访问该应用程序。为什么服务在正常启动时不起作用，并且仅在调试模式下工作。

已经阅读了有关类似错误的多个帖子，但没有任何帮助。我正在使用Python 3.8，并将VENV脚本目录也添加到路径中。有人可以帮忙吗？

更新 使用我的用户帐户开始服务，并且它可以正常工作，但仍然不了解其在调试模式下的工作方式。