解决python写的windows服务不能启动的问题

364次阅读  |  发布于5年以前

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"}],
)

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