如何为继承的重载方法起别名？

Question

如何从基类使用重命名的重载方法并保留输出签名？

方法必须在继承自基类的类中有一个新名称

例如：

class BaseApi(api_client.Api):
    @typing.overload
    def _redirection(
        self,
        server_index: typing.Optional[int] = None,
        stream: bool = False,
        timeout: typing.Optional[typing.Union[int, typing.Tuple]] = None,
        skip_deserialization: typing_extensions.Literal[False] = ...,
    ) -> typing.Union[
        response_3xx.ResponseFor3XX.response_cls,
        response_303.ResponseFor303.response_cls,
    ]: ...

    @typing.overload
    def _redirection(
        self,
        skip_deserialization: typing_extensions.Literal[True],
        server_index: typing.Optional[int] = None,
        stream: bool = False,
        timeout: typing.Optional[typing.Union[int, typing.Tuple]] = None,
    ) -> api_client.ApiResponseWithoutDeserialization: ...

    @typing.overload
    def _redirection(
        self,
        server_index: typing.Optional[int] = None,
        stream: bool = False,
        timeout: typing.Optional[typing.Union[int, typing.Tuple]] = None,
        skip_deserialization: bool = ...,
    ) -> typing.Union[
        response_3xx.ResponseFor3XX.response_cls,
        response_303.ResponseFor303.response_cls,
        api_client.ApiResponseWithoutDeserialization,
    ]: ...

    def _redirection(
        self,
        server_index: typing.Optional[int] = None,
        stream: bool = False,
        timeout: typing.Optional[typing.Union[int, typing.Tuple]] = None,
        skip_deserialization: bool = False,
    ):
        # implementation

class ApiForGet(BaseApi):
    # this class is used by api classes that refer to endpoints by path and http method names

    def get(
        self,
        server_index: typing.Optional[int] = None,
        stream: bool = False,
        timeout: typing.Optional[typing.Union[int, typing.Tuple]] = None,
        skip_deserialization: bool = False,
    ):
        return self._redirection(
            server_index=server_index,
            stream=stream,
            timeout=timeout,
            skip_deserialization=skip_deserialization
        )

class Redirection(BaseApi):
    # this class is used by api classes that refer to endpoints with operationId.snakeCase fn names

    def redirection(
        self,
        server_index: typing.Optional[int] = None,
        stream: bool = False,
        timeout: typing.Optional[typing.Union[int, typing.Tuple]] = None,
        skip_deserialization: bool = False,
    ):
        return self._redirection(
            server_index=server_index,
            stream=stream,
            timeout=timeout,
            skip_deserialization=skip_deserialization
        )

当我打电话

res = BaseApi._redirection(skip_deserialization=True)

我明白了

ApiResponseWithoutDeserialiation

但是当我使用时：

res = ApiForGet._redirection(skip_deserialization=True)

我明白了

ApiResponseFor3XX | ApiResponseFor3XX | ApiResponseWithoutDeserialiation

注：我用的是vscode

我能想到的选项是：

• 在 ApiForGet + Redirection 中编写重载，但这会复制它们 2X
• 使用某种装饰魔法重命名方法和重载方法

我该怎么做？ 为什么我输入的内容没有传递到更深层次的调用站点？

这些工作，还有更好的方法吗？

class ApiForGet(BaseApi):
    # this class is used by api classes that refer to endpoints by path and http method names

    get = BaseApi._redirection


class ApiForGet(BaseApi):
    pass

ApiForGet.get = ApiForGet._redirection

Answer 1

这些工作：

class ApiForGet(BaseApi):
    # this class is used by api classes that refer to endpoints by path and http method names

    get = BaseApi._redirection


class ApiForGet(BaseApi):
    pass

ApiForGet.get = ApiForGet._redirection