Typescript: define type like keyof but with specific value type

Question

I have interface

interface MyInterface {
  field1: boolean,
  field2: MyType,
  field3: MyType
}

and I want to create type that contains keys of this interface but only those which usage in interface gives value with type MyType. I know about existence of keyof but it will return ALL keys even field1 which I don't need. So how can I get type with only field2 and field3?

Aleksey L. · Accepted Answer · 2020-04-02 07:06:03Z

1

You could create mapped type that checks if value extends MyType and if yes takes the key otherwise puts never, then index into it with all possible keys (to produce a union of keys that their values extend MyType):

type PickKeysOfType<T, TValue> = {
    [P in keyof T]: T[P] extends TValue ? P : never
}[keyof T];

type MyTypeKeys = PickKeysOfType<MyInterface, MyType> // "field2" | "field3"

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answered Apr 2, 2020 at 7:06

Aleksey L.

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