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：
#!/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"}],
)
![详解Python中 sys.argv[]的用法简明解释](/zb_users/upload/202003/jjzmf5btigc.png)
