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\”}],
)