typescript - 有没有办法以数字形式获取元组的键？

Question

考虑我有一些元组类型:

type SomeTuple = ["a", "b", "c"];

现在我想以某种方式获取它的键作为数字:

type SomeTupleKeys<T = SomeTuple> = 0 | 1 | 2;

我尝试使用keyof SomeTuple，但它返回

number | "0" | "1" | "2" | "length" | "toString" | "toLocaleString" | "pop" | "push" | "concat" | "join" | "reverse" | "shift" | "slice" | "sort" | "splice" | "unshift" | "indexOf" | ... 14 more ... | "values"

我也尝试过更复杂的解决方案

type B = (SomeTuple extends infer T ? { [K in keyof T]: K extends keyof T ? K : never } : never)[number];

但是这个给了我字符串:

"0" | "1" | "2"

有没有一种方法可以在不硬编码任何内容的情况下获取数字？

Demo

Answer 1

这可以通过 recursive conditional types 来完成(在 Typescript 4.1 中添加):

type TupleKeys<T extends ReadonlyArray<unknown>, Keys extends ReadonlyArray<number> = []> =
    T extends [infer _, ... infer R]
        ? TupleKeys<R, [R['length'], ...Keys]>
        : Keys[number];

type SomeTuple = ["a", "b", "c"];
type SomeTupleKeys = TupleKeys<SomeTuple>; // 0 | 2 | 1

Playground

---

另一个更简单的解决方案(使用较旧的 typescript 版本):

type TupleKeys<T extends Array<any>> = Exclude<Partial<T>['length'], T['length']>;

type SomeTuple = ["a", "b", "c"];
type SomeTupleKeys = TupleKeys<SomeTuple>; // 0 | 1 | 2

这里，通过使用Partial，我们得到长度为0或1或2或3的元组的并集，然后我们排除3(原始元组的长度)

Playground