Typescript 如何键入对象转换器

发布于11月10日

我有以下目的:

const schema = {
  student: {id: 'student/student'},
  students: {id: 'student/students'},
} as const;

interface Schema {
    readonly [key: string]: {id: string};
}

type Mapped<V extends Schema, U extends keyof V> = {
    [S in V[U]['id']]: V[U];
}

但是当我使用这种类型时，所有的值都可以在每个键上使用

const a: Mapped<typeof schema, keyof typeof schema> = {
    'student/student': {id: 'student/student'},
    'student/students': {id: 'student/student'},
}

推荐答案

原因

type Mapped<V extends Schema, U extends keyof V> = {
    [S in V[U]['id']]: V[U];
}

不起作用是因为union type-V[U]独立于S.您需要提取与S对应的V[U]的成员才能正常工作:

type Mapped<V extends Schema, U extends keyof V> = {
  [S in V[U]['id']]: Extract<V[U], { id: S }>;
}

type M = Mapped<typeof schema, keyof typeof schema>;
/* type M = {
    "student/student": {
        readonly id: "student/student";
    };
    "student/students": {
        readonly id: "student/students";
    };
} */

在这里，我使用Extract utility type将V[U]筛选为仅匹配{id: S}的成员.

一个更简单的方法是直接覆盖V[U]的union成员，然后remap这些成员的id属性的键:

type Mapped<V extends Schema, U extends keyof V> = {
  [W in V[U] as W["id"]]: W;
}

type M = Mapped<typeof schema, keyof typeof schema>;
/* type M = {
    "student/student": {
        readonly id: "student/student";
    };
    "student/students": {
        readonly id: "student/students";
    };
} */

因此，您不必丢弃id属性以外的所有属性S，然后在V[U]中重新搜索以找到合适的unions 成员，而是始终持有合适的unions 成员W.

Playground link to code个