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/