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    解决python写的windows服务不能启动的问题

    作者:admin 时间:2021-07-04 17:43

    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"}],
    )

     

    jsjbwy