如何将字符串联合类型中的所有字符串组成一个数组？

Question

type stringUnion = "Value1"|"Value2" 我想从所有字符串中获取并创建一个数组stringUnion

所以数组看起来像这样["Value1","Value2"]，我想要一个数组，这样我就可以像这样迭代它：

theArray.map((current, index)=>{

})

我怎么做？

Answer 1

对的，这是可能的。

// credits goes to https://stackoverflow.com/questions/50374908/transform-union-type-to-intersection-type/50375286#50375286
type UnionToIntersection<U> = (U extends any ? (k: U) => void : never) extends (
  k: infer I
) => void
  ? I
  : never;

// credits goes to https://github.com/microsoft/TypeScript/issues/13298#issuecomment-468114901
type UnionToOvlds<U> = UnionToIntersection<
  U extends any ? (f: U) => void : never
>;

type PopUnion<U> = UnionToOvlds<U> extends (a: infer A) => void ? A : never;


// credits goes to https://stackoverflow.com/questions/53953814/typescript-check-if-a-type-is-a-union#comment-94748994
type IsUnion<T> = [T] extends [UnionToIntersection<T>] ? false : true;

type UnionToArray<T, A extends unknown[] = []> = IsUnion<T> extends true
  ? UnionToArray<Exclude<T, PopUnion<T>>, [PopUnion<T>, ...A]>
  : [T, ...A];

interface Person {
  name: string;
  age: number;
  surname: string;
  children: number;
}

type Result = UnionToArray<keyof Person>; // ["name", "age", "surname", "children"]

这是另一种方法：

// credits goes to https://twitter.com/WrocTypeScript/status/1306296710407352321
type TupleUnion<U extends string, R extends any[] = []> = {
    [S in U]: Exclude<U, S> extends never ? [...R, S] : TupleUnion<Exclude<U, S>, [...R, S]>;
}[U];

interface Person {
    firstName: string;
    lastName: string;
    dob: Date;
    hasCats: false;
}

type keys = TupleUnion<keyof Person>;

操场

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