我已经用 python 编写了一个 Windows 服务。如果我从命令提示符运行我的脚本

python runService.py 

当我这样做时，服务会正确安装并启动。我一直在尝试使用 pyinstaller 创建可执行文件，因为我在 py2exe 上看到了同样的问题。当我运行 .exe 时，服务会安装但未启动，并且出现以下错误

error 1053 the service did not respond to the start or control request in a timely fashion 

我看到很多人都遇到过这个问题，但我似乎找不到关于如何解决这个问题的明确答案。

winservice.py

from os.path import splitext, abspath
from sys import modules, executable
from time import *
import win32serviceutil
import win32service
import win32event
import win32api

class Service(win32serviceutil.ServiceFramework):
    _svc_name_ = '_unNamed'
    _svc_display_name_ = '_Service Template'
    _svc_description_ = '_Description template'
    def __init__(self, *args):
        win32serviceutil.ServiceFramework.__init__(self, *args)
        self.log('init')
        self.stop_event = win32event.CreateEvent(None, 0, 0, None)

    #logs into the system event log
def log(self, msg):
    import servicemanager
    servicemanager.LogInfoMsg(str(msg))

def sleep(self, minute):
        win32api.Sleep((minute*1000), True)
def SvcDoRun(self):
    self.ReportServiceStatus(win32service.SERVICE_START_PENDING)
    try:
        self.ReportServiceStatus(win32service.SERVICE_RUNNING)
        self.log('start')
        self.start()
        self.log('wait')
        win32event.WaitForSingleObject(self.stop_event, win32event.INFINITE)
        self.log('done')
    except Exception, x:
        self.log('Exception : %s' % x)
        self.SvcStop()
def SvcStop(self):
    self.ReportServiceStatus(win32service.SERVICE_STOP_PENDING)
    #self.log('stopping')
    self.stop()
    #self.log('stopped')
    win32event.SetEvent(self.stop_event)
    self.ReportServiceStatus(win32service.SERVICE_STOPPED)
# to be overridden
def start(self): pass
# to be overridden
def stop(self): pass

def instart(cls, name, description, display_name=None, stay_alive=True):
    ''' Install and  Start (auto) a Service

        cls : the class (derived from Service) that implement the Service
        name : Service name
        display_name : the name displayed in the service manager
        decription: the description 
        stay_alive : Service will stop on logout if False
    '''
    cls._svc_name_ = name
    cls._svc_display_name_ = display_name or name
    cls._svc_desciption_ = description
    try:
        module_path=modules[cls.__module__].__file__
    except AttributeError:

        module_path=executable
    module_file = splitext(abspath(module_path))[0]
    cls._svc_reg_class_ = '%s.%s' % (module_file, cls.__name__)
    if stay_alive: win32api.SetConsoleCtrlHandler(lambda x: True, True)
    try:
        win32serviceutil.InstallService(
            cls._svc_reg_class_,
            cls._svc_name_,
            cls._svc_display_name_,
            startType = win32service.SERVICE_AUTO_START,
            description = cls._svc_desciption_
        )
        print 'Install ok'
        win32serviceutil.StartService(
        cls._svc_name_
    )
    print 'Start ok'
except Exception, x:
    print str(x)

更新

我通过使用 py2exe 解决了这个问题，但相同的更改也可能适用于 pyinstaller。我没有时间亲自检查一下。

我不得不删除 instart 功能。以下是我的 winservice.py 现在的读取方式。

winservice_py2exe.py

from os.path import splitext, abspath
from sys import modules, executable
from time import *
import win32serviceutil
import win32service
import win32event
import win32api

class Service(win32serviceutil.ServiceFramework):
    _svc_name_ = 'actualServiceName' #here is now the name you would input as an arg for instart
    _svc_display_name_ = 'actualDisplayName' #arg for instart
    _svc_description_ = 'actualDescription'# arg from instart
    def __init__(self, *args):
        win32serviceutil.ServiceFramework.__init__(self, *args)
        self.log('init')
        self.stop_event = win32event.CreateEvent(None, 0, 0, None)

    #logs into the system event log
def log(self, msg):
    import servicemanager
    servicemanager.LogInfoMsg(str(msg))

def sleep(self, minute):
        win32api.Sleep((minute*1000), True)
def SvcDoRun(self):
    self.ReportServiceStatus(win32service.SERVICE_START_PENDING)
    try:
        self.ReportServiceStatus(win32service.SERVICE_RUNNING)
        self.log('start')
        self.start()
        self.log('wait')
        win32event.WaitForSingleObject(self.stop_event, win32event.INFINITE)
        self.log('done')
    except Exception, x:
        self.log('Exception : %s' % x)
        self.SvcStop()
def SvcStop(self):
    self.ReportServiceStatus(win32service.SERVICE_STOP_PENDING)
    #self.log('stopping')
    self.stop()
    #self.log('stopped')
    win32event.SetEvent(self.stop_event)
    self.ReportServiceStatus(win32service.SERVICE_STOPPED)
# to be overridden
def start(self): pass
# to be overridden
def stop(self): pass

if __name__ == '__main__':
     # Note that this code will not be run in the 'frozen' exe-file!!!
     win32serviceutil.HandleCommandLine(VidiagService) #added from example included with py2exe

下面是我在 py2exe 中使用的 setup.py 文件。这取自 py2exe 安装中包含的示例:

setup.py

from distutils.core import setup
import py2exe
import sys
if len(sys.argv) == 1:
    sys.argv.append("py2exe")
    sys.argv.append("-q")

class Target:
    def __init__(self, **kw):
    self.__dict__.update(kw)
     # for the versioninfo resources
     self.version = "0.5.0"
     self.company_name = "No Company"
     self.copyright = "no copyright"
     self.name = "py2exe sample files"

 myservice = Target(
     # used for the versioninfo resource
     description = "A sample Windows NT service",
     # what to build.  For a service, the module name (not the
     # filename) must be specified!
     modules = ["winservice_py2exe"]
     )

 setup(
     options = {"py2exe": {"typelibs":
                      # typelib for WMI
                       [('{565783C6-CB41-11D1-8B02-00600806D9B6}', 0, 1, 2)],
                       # create a compressed zip archive
                       "compressed": 1,
                       "optimize": 2}},
     # The lib directory contains everything except the executables and the python dll.
     # Can include a subdirectory name.
     zipfile = "lib/shared.zip",

     service = [myservice]
    )

创建 exe 后，您可以使用以下命令从命令安装服务

winservice_py2exe.exe -install

然后启动你可以使用的服务:

 net start aTest

或来自 Windows 服务管理器。所有其他 Windows 命令行功能现在都可以在该服务以及 Windows 服务管理器中使用。

最佳答案

尝试将最后几行更改为

if __name__ == '__main__':
    if len(sys.argv) == 1:
        servicemanager.Initialize()
        servicemanager.PrepareToHostSingle(Service)
        servicemanager.StartServiceCtrlDispatcher()
    else:
        win32serviceutil.HandleCommandLine(Service)

关于Python Windows 服务 pyinstaller 可执行文件错误 1053，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/25770873/