我有这个代码：
from os.path import splitext, abspath from sys import modules import win32serviceutil import win32service import win32event import win32api def main(): # do something class Service(win32serviceutil.ServiceFramework): _svc_name_ = '_unNamed' _svc_display_name_ = '_Service Template' def __init__(self, *args): win32serviceutil.ServiceFramework.__init__(self, *args) self.log('init') self.stop_event = win32event.CreateEvent(None, 0, 0, None) def log(self, msg): import servicemanager servicemanager.LogInfoMsg(str(msg)) def sleep(self, sec): win32api.Sleep(sec*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 as 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, 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 stay_alive : Service will stop on logout if False ''' cls._svc_name_ = name cls._svc_display_name_ = display_name or name try: module_path=modules[cls.__module__].__file__ except AttributeError: # maybe py2exe went by from sys import executable 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 ) print ('Install ok') win32serviceutil.StartService( cls._svc_name_ ) print ('Start ok') except Exception as x: print (str(x)) startmain() # to start the main function of the program Service = Service instart = instart class Test(Service): def start(self): self.runflag=True while self.runflag: self.sleep(10) self.log("I'm alive ...") def stop(self): self.runflag=False self.log("I'm done") instart(Test, 'aTest', 'Python Service Test') 该服务已成功创build并启动，但具有pipe理员权限。 如果我以普通用户身份执行脚本，则失败。 有没有pipe理员权限做到这一点？ 我还注意到，这个类不仅启动脚本作为服务，但它也创build一个Services.msc。 但我不需要它。 我可以把我的脚本自动运行（这是没有pipe理员烫发可行）。重要的是，我的脚本已经开始在后台。 我可以这样做吗？