TypeScript:从有区别的联合派生地图 [英] TypeScript: derive map from discriminated union

问题描述

我有一个区分的联合类型，该类型根据字符串文字字段来区分类型.我想派生一个映射类型，该映射类型将并集中的所有类型映射到其相应的鉴别符文字值.

I have a discriminated union type that differentiates types based on a string literal field. I would like to derive a mapped type that maps all of the types in the union to their corresponding discriminator literal values.

例如

export type Fetch = {
    type: 'fetch',
    dataType: string
};

export type Fetched<T> = {
    type: 'fetched',
    value: T
};

// union type discriminated on 'type' property
export type Action =
    | Fetch
    | Fetched<Product>;

// This produces a type 'fetch' | 'fetched'
// from the type 
type Actions = Action['type'];

// I want to produce a map type of the discriminator values to the types 
// comprising the union type but in an automated fashion similar to how I
// derived my Actions type.
// e.g.
type WhatIWant = {
    fetch: Fetch,
    fetched: Fetched<Product>
}

在TypeScript中可以吗?

Is this possible in TypeScript?

推荐答案

随着

With the introduction of conditional types in TypeScript 2.8, you can define a type function which, given a discriminated union and the key and value of the discriminant, produces the single relevant constituent of the union:

type DiscriminateUnion<T, K extends keyof T, V extends T[K]> = 
  T extends Record<K, V> ? T : never

如果您想使用它来构建地图，也可以这样做:

And if you wanted to use that to build up a map, you can do that too:

type MapDiscriminatedUnion<T extends Record<K, string>, K extends keyof T> =
  { [V in T[K]]: DiscriminateUnion<T, K, V> };

所以在您的情况下，

type WhatIWant = MapDiscriminatedUnion<Action, 'type'>;

如果您对其进行检查，则为:

which, if you inspect it, is:

type WhatIWant = {
  fetch: {
    type: "fetch";
    dataType: string;
  };
  fetched: {
    type: "fetched";
    value: Product;
  };
}

我认为是所希望的.希望能有所帮助；祝你好运！

as desired, I think. Hope that helps; good luck!

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