获取按 object 值类型过滤的 object 键的联合类型

Question

Is there a way to get union type of object keys filtered by type of object value?

I think, type inferring in conditional type is suitable for that, so I tried below. ( KeyOfValueType is what I want)

type Obj1 = { a: string, b: number };
type Obj2 = { c: number, d: string, e: string };

type KeyOfValueType<
    Obj extends {},
    ValueType extends Obj[keyof Obj]
> = ValueType extends Obj[infer K] ? K : never;

type Result1 = KeyOfValueType<Obj1, number>; // "b" because Obj1["b"] is equal to number
type Result2 = KeyOfValueType<Obj2, string>; // "d" | "e" because Obj2["d"] and Obj2["e"] are equals to string;

TypeScript Playground

But tt raises an error Type 'K' cannot be used to index type 'Obj'. at Obj[infer K] .

Answer 1

There is the package https://github.com/piotrwitek/utility-types . It has PickByValue and PickByValueExact operators(and many other useful type operators). You can use them along with keyof to achieve your goal. They indeed use conditional type and infer for that.

type PickByValue<T, ValueType> = Pick<T, {
    [Key in keyof T]: T[Key] extends ValueType ? Key : never;
}[keyof T]>;

type R1 = keyof PickByValue<Obj1, number> //b
type R2 = keyof PickByValue<Obj2, string> //d | e