如何告诉 Typescript 检查 object[key] 是否属于某种类型？

Question

假设我有一个这样的 Typescript 函数：

const myFunction = <U, V extends keyof U> (object: U, key: V) => {
    // ...
}

如何让 Typescript 检查 object[key] 是否属于某种类型？

例如，假设我希望 object[key] 为“字符串”类型。

所以，当我打电话给myFunction 时，它应该是这样工作的：

myFunction({keyA: "hello", keyB: 123}, "keyA") // OK

myFunction({keyA: "hello", keyB: 123}, "keyB") // error: `keyB` must be of type string but is number

Answer 1

根据In TypeScript, how to get the keys of an object type whose values are of a given type? 的答案，我整理了这个工作示例，专门回答了您的问题（see playground），但困难的部分确实来自另一个答案。

type KeysMatching<T,V> = keyof { [ P in keyof T as T[P] extends V ? P : never ] : P };

function myFunction<T>(object: T, key: KeysMatching<T,string> ){
    // ...
}

myFunction({keyA: "hello", keyB: 123}, "keyA") // OK

myFunction({keyA: "hello", keyB: 123}, "keyB") // error: `keyB` must be of type string but is number

---

我非常喜欢提取 PickValues<T,V> 类型的形式，它将 T 的属性过滤为仅具有 V 中的值的属性。然后只需在上面使用 keyof...

type PickValues<T,V> = { [ P in keyof T as T[P] extends V ? P : never ] : T[P] }
type Picked = PickValues<{keyA: "hello", keyB: 123},string>;
// type Picked = {
//     keyA: "hello";
// }

//Your case
function myFunction<T>(object: T, key: keyof PickValues<T,string> ){
    // ...
}
myFunction({keyA: "hello", keyB: 123}, "keyA") // OK
myFunction({keyA: "hello", keyB: 123}, "keyB") // error: `keyB` must be of type string but is number

顺便说一句，这个策略可以被推广到做相反的事情并创建一个OmitValues<T,V> 类型......

type OmitValues<T,V> = { [ P in keyof T as T[P] extends V ? never : P ] : T[P] }
type Omitted = OmitValues<{keyA: "hello", keyB: 123},string>;
// type Omitted = {
//     keyB: 123;
// }

最后，它也可以重新用于仅提取具有可选性质的属性...

type OptionalValues<T> = { [ P in keyof T as undefined extends T[P] ? P : never ] : T[P] }
type Optional = OptionalValues<{keyA?: "hello", keyB: 123}>;
// type Optional = {
//     keyA?: "hello" | undefined;
// }