是否有可能做這樣的事情：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" 我想創建一個類，其他類可以從中繼承并根據類名設置其名稱。在python 3中可能嗎？
2 回答
飲歌長嘯
TA貢獻1951條經驗 獲得超3個贊
你不需要它，它已經在那里了：
>>> class Parent(): pass
...
>>> class Child(): pass
...
>>> Parent.__name__
'Parent'
>>> Child.__name__
'Child'
>>>
但是，如果您真的出于不費心解釋的任何原因堅持自己滾動，則可以使用自定義元類：
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
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