解决python写的windows服务不能启动的问题

报“服务没有及时响应或控制请求”的错误，改用pyinstaller生成也是不行；查资料后修改setup.py如下即可,服务名、脚本名请自行替换：

复制代码 代码如下:
#!/usr/bin/python 
#-*-coding:cp936-*-
from distutils.core import setup
import py2exe

class Target:
    def __init__(self, **kw):
        self.__dict__.update(kw)
        # for the versioninfo resources
        self.version = \”1.1.8\”
        self.company_name = \”Yovole Shanghai Co. Ltd.\”
        self.copyright = \”Copyright (c) 2013 Founder Software (Shanghai) Co., Ltd. \”
        self.name = \”Guest Agent\”

myservice = Target(
    description = \’Yovole Cloud Desktop Guest Agent\’,
    modules = [\’service\’],
    cmdline_style=\’pywin32\’
    #icon_resources=[(1, \”cartrigde.ico\”)]
)

options = {\”py2exe\”:  
            {   \”compressed\”: 1,  
                \”bundle_files\”: 1
            }  
          } 

setup(
    service=[myservice],
    options = options,
    zipfile = None,
    windows=[{\”script\”: \”service.py\”}],
)