typescript - 映射类型的交集

Question

考虑以下几点:

type Properties = {
    foo: { n: number };
    bar: { s: string };
    baz: { b: boolean };
};

declare function retrieveValues<K extends keyof Properties>(add?: K[]): Pick<Properties, K>[K];

// what happens
const x: { n: number } | { s: string } = retrieveValues(['foo', 'bar']);

// what I'm really trying to express (type error)
const y: { n: number } & { s: string } = retrieveValues(['foo', 'bar']);

有没有办法得到Pick<Properties, K>的属性的交集？ ？或者只是根据数组中相关字符串的存在来获取一组类型的交集的不同方法？

Answer 1

使用 conditional types和 type inference in conditional types可以转换 { n: number } | { s: string } 直接进入 { n: number } & { s: string } 。

type GetKeys<U> = U extends Record<infer K, any> ? K : never

type UnionToIntersection<U extends object> = {
   [K in GetKeys<U>]: U extends Record<K, infer T> ? T : never
}

type Transformed = UnionToIntersection<{ a: string } | { b: number }>
// Transformed has type {a: string, b: number}

Playground Link

之所以可行，主要是因为条件类型分布在联合类型上。来自conditional types pull request :

Conditional types in which the checked type is a naked type parameter are called distributive conditional types. Distributive conditional types are automatically distributed over union types during instantiation. For example, an instantiation of T extends U ? X : Y with the type argument A | B | C for T is resolved as (A extends U ? X : Y) | (B extends U ? X : Y) | (C extends U ? X : Y).