解决python写的windows服务不能启动的问题_python

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"}],
)