在子类型中使用索引中的类型

Question

我有一堆内置在对象文字中的数据。这个对象的键也存在于这个对象的每个子对象中，我想用打字稿来强制执行：

type People = {[NAME in string]: { name: NAME; age: number }};

const people: People = {
    Jeff: { name: "Jeff", age: 30 },
    Fred: { name: "Freddy", age: 25 }, // no error, name evaluates to string
    Sally: { name: "Sally", age: 35 },
};

过去我使用字符串联合做过类似的事情：

type People = {[NAME in "Jeff" | "Fred" | "Sally"]: { name: NAME; age: number }};

const people: People = {
    Jeff: { name: "Jeff", age: 30 },
    Fred: { name: "Freddy", age: 25 }, // ERROR: Type '"Freddy"' is not assignable to type '"Fred"'.  ts(2322)
    Sally: { name: "Sally", age: 35 },
};

但是对于我当前的用例来说，除了对象之外维护一个单独的联合类型是多余的。

是否可以在该对象的子对象中使用索引对象的键类型？

Answer 1

我能做的最好是这样：

type People<T> = { [K in keyof T]: { name: K; age: number } };

function createPeople<T extends People<T>>(people: T): People<T> {
    return people;
}

const people = createPeople({
    Jeff: { name: 'Jeff', age: 30 },
    Fred: { name: 'Freddy', age: 25 }, // ERROR: Type '"Freddy"' is not assignable to type '"Fred"'.  ts(2322)
    Sally: { name: 'Sally', age: 35 },
});