TypeScript - 根据鉴别器将联合类型映射到另一个联合类型的函数

Question

我有 2 个使用相同鉴别器字段 + 值的可区分联合类型。我正在尝试编写一个函数，可以根据鉴别器将 1 映射到另一个。

例如

输入类型：

type InA = {
    type: 'a',
    data: string
};

type InB = {
    type: 'b',
    data: number
};

type In = InA | InB;

输出类型：

type OutA = {
    type: 'a',
    data: Object
};

type OutB = {
    type: 'b',
    data: Array<number>
};

type Out = OutA | OutB;

映射功能

// This is the function I'd like to have a better type signature
// for inferring output type based on input type
function map<In, Out>(
    in: In
): Out {
   // do something
}

用法

// I want the compiler to infer that this is OutB based on the InB
let result = map({ type: 'b', value: 999 });

有没有办法编写映射的函数签名以便这可以工作？

回答后更新 我能够使用 @titian-cernicova-dragomir 答案的修改版本。这是总体思路+一些关于我如何使用它的附加上下文：

/** Http Request types **/

type RequestA = {
    type: 'names',
    url: '/names'
};

type RequestB = {
    type: 'numbers',
    url: '/numbers'
};

type Request = RequestA | RequestB;

/** Response types **/

type ResponseA = {
    type: 'names',
    data: Array<string>
};

type ResponseB = {
    type: 'numbers',
    data: Array<number>
};

type Response = ResponseA | ResponseB;

/** Helper from accepted answer */
type GetOut<T, A> = T extends { type: A } ? T : never;

/** Generic function for fetching data */
export function fetchData<
    Req extends Request,
    Res extends GetOut<Response, Req['type']>
    >(request: Req): Promise<Res> {
    return fetch(request.url)
        .then(response => response.json())
        .then(data => {
            return <Res>{
                type: request.type,
                data
            }
        });
}

// compiler knows that this is of type Promise<ResponseA> based on
// type discriminiator
let names = fetchData({
    type: 'names',
    url: '/names'
});

// compiler knows that this is of type Promise<ResponseB> based on
// type discriminiator
let numbers = fetchData({
    type: 'numbers',
    url: '/numbers'
});

Answer 1

您可以使用条件类型来执行此操作。首先，我们使用条件类型来提取作为参数传递的实际字符串文字类型（我们将其称为A）。然后使用A我们将过滤Out以从扩展的 unon 中获取类型{ type: A }。

type GetOut<T, A> = T extends { type : A} ? T: never; 
function map2<TIn extends In>(inParam: TIn) : TIn extends { type: infer A } ? GetOut<Out, A> : never {
    return null as any;
}

let resultA = map2({ type: 'a', data: '999' }); // result is OutA
let resultB = map2({ type: 'b', data: 999 }); // result is OutB

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