Question

如何将正确的类型关联函数参数作为元组返回？我想要一个函数根据其参数的元组类型返回其值。

下面的例子是一个 get 函数，但我想让它使用 getAll 函数和参数作为元组。

type PathImpl<T, Key extends keyof T> =
  Key extends string
  ? T[Key] extends Record<string, any>
    ? | `${Key}.${PathImpl<T[Key], Exclude<keyof T[Key], keyof any[]>> & string}`
      | `${Key}.${Exclude<keyof T[Key], keyof any[]> & string}`
    : never
  : never;

type PathImpl2<T> = PathImpl<T, keyof T> | keyof T;

type Path<T> = PathImpl2<T> extends string | keyof T ? PathImpl2<T> : keyof T;

type PathValue<T, P extends Path<T>> =
  P extends `${infer Key}.${infer Rest}`
  ? Key extends keyof T
    ? Rest extends Path<T[Key]>
      ? PathValue<T[Key], Rest>
      : never
    : never
  : P extends keyof T
    ? T[P]
    : never;

declare function get<T, P extends Path<T>>(obj: T, path: P): PathValue<T, P>;

const object = {
  firstName: "test",
  lastName: "test1"
} as const;

get(object, "firstName");

declare function getAll<T, P extends Path<T>>(obj: T, args: P[]): PathValue<T, P>;

const data = getAll(object, ['firstName', 'lastName'])
// how to produce the type: ['test', 'test1']

Playground

Answer 1

您可以使用映射类型来映射传入的元组。您还必须更改类型参数以捕获元组类型。

type MapAllPaths<T, P extends Path<T>[]> = {} & {
  [K in keyof P ]: PathValue<T, P[K] & Path<T>>
}

declare function getAll<T, P extends [Path<T>] | Path<T>[]>(obj: T, args: P): MapAllPaths<T, P>;

Playground Link

具有函数正确返回类型的元组参数

1 个答案: