如何在Typescript 3.0中将元组“映射"到另一个元组类型 [英] How to 'map' a Tuple to another Tuple type in Typescript 3.0

问题描述

我有Maybe类型的元组:

class Maybe<T>{ }

type MaybeTuple = [Maybe<string>, Maybe<number>, Maybe<boolean>];

我想把它变成一个类型的元组:

and I want to turn this into a tuple of types:

type TupleIWant = [string, number, boolean];

所以我尝试了这个:

type ExtractTypes<T> = T extends Maybe<infer MaybeTypes>[] ? MaybeTypes : never;

type TypesArray = ExtractTypes<MaybeTuple>; // string | number | boolean NOT [string, number, boolean]

不起作用的地方:-(

我得到(string | number | boolean)[]而不是我想要的元组:[string, number, boolean]

I get (string | number | boolean)[] rather than the tuple I want: [string, number, boolean]

我现在想做什么?

推荐答案

您将需要使用

You'll need to use a mapped tuple type, which is supported in TypeScript 3.1.

您可以创建具有正确类型的属性0，1，2和length的映射类型，如下所示:

You can make a mapped type that has properties 0, 1, 2 and length of the correct types, like this:

class Maybe<T> {}

type MaybeTuple = [Maybe<string>, Maybe<number>, Maybe<boolean>];

type MaybeType<T> = T extends Maybe<infer MaybeType> ? MaybeType : never;
type MaybeTypes<Tuple extends [...any[]]> = {
  [Index in keyof Tuple]: MaybeType<Tuple[Index]>;
} & {length: Tuple['length']};

let extractedTypes: MaybeTypes<MaybeTuple> = ['hello', 3, true];

如果您使用的是较早版本的Typescript，则可以使用TypeScript的未发布版本，或者作为一种变通方法，可以编写条件类型以与元组匹配，只要您认为自己可能会拥有，例如，像这样.

If you're using an older version of typescript, you can use an unreleased version of TypeScript, or as a workaround, you can write a conditional type to match against tuples as long as you think you are likely to have, e.g., like this.

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