继承后设置类属性

Question

是否有可能做这样的事情：

class ParentClassName(object):

    name = camel_case_to_underscore(cls.__name__)

class ChildClassName(Parent):

    pass

assert Parent().name == "parent_class_name" 

assert Child().name == "child_class_name" 

assert Child.name == "child_class_name" 

assert getattr(Child, 'name') == "child_class_name" 

我想创建一个类，其他类可以从中继承并根据类名设置其名称。在python 3中可能吗？

Answer 1

你不需要它，它已经在那里了：>>> class Parent(): pass... >>> class Child(): pass... >>> Parent.__name__'Parent'>>> Child.__name__'Child'>>> 但是，如果您真的出于不费心解释的任何原因坚持自己滚动，则可以使用自定义元类：def camel_case_to_underscore(name):    # your code hereclass NamedType(type):    def __new__(meta, name, bases, attribs):        attribs["name"] = camel_case_to_underscore(name)        return type.__new__(meta, name, bases, attribs)class Parent(metaclass=NamedType):    passclass Child(Parent):    pass