python 父子属性访问？

Question

from loguru import logger


class State:
    def __init__(self):
        self.logger = logger

    def __setattr__(self, name, value):
        self.logger.info(f'wwwwasdasd')# 问题代码
        self.__dict__[name] = value


class LocalState(State):
    def __init__(self):
        super().__init__()
        self.ll = 2


if __name__ == '__main__':
    state = LocalState()
    state.ll = 22222
    print(state.ll)

怎么可以让这个 self.logger 属于子类（不是直接添加到子类属性这种方式），并且可以运行

Answer 1

from loguru import logger


class State:
    def __setattr__(self, name, value):
        if hasattr(self, 'logger'):
            self.logger.info(f'wwwwasdasd')
        super().__setattr__(name, value)  # 使用 super() 委托给父类


class LocalState(State):
    def __init__(self):
        super().__init__()
        self.logger = logger
        self.ll = 2


if __name__ == '__main__':
    state = LocalState()
    state.ll = 22222
    print(state.ll)
from loguru import logger


class State:
    def __setattr__(self, name, value):
        if hasattr(self, 'logger'):
            self.logger.info(f'wwwwasdasd')
        super().__setattr__(name, value)  # 使用 super() 委托给父类


class LocalState(State):
    def __init__(self):
        super().__init__()
        self.logger = logger
        self.ll = 2


if __name__ == '__main__':
    state = LocalState()
    state.ll = 22222
    print(state.ll)

Answer 2

import abc

from loguru import logger


class State:
    def __init__(self):
        pass

    @property
    @abc.abstractmethod
    def get_logger(self):
        raise "子类实现该方法"

    def __setattr__(self, name, value):
        self.get_logger.info(f"wwwwasdasd")  # 问题代码
        self.__dict__[name] = value


class LocalState(State):
    def __init__(self):
        super().__init__()
        self.ll = 2

    @property
    def get_logger(self):
        # 可以返回任意一个logger，父类调用该方法取决于子类的实现
        return logger


if __name__ == "__main__":
    state = LocalState()
    state.ll = 22222
    print(state.ll)