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):    passassert 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 here

class 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):

    pass

class Child(Parent):

    pass