Pick type by array of string

Question

I have type

type User = {
    firstName: string,
    lastName: string,
    age: string
}

I need a function

const result =  fun<User>(['firstName', 'lastName'])

I need result type is a User with picked firstName and lastName. Is it possible to do with typeScript?

Answer 1

Yes, it is possible. I did not came up with this solution myself so all credits go to "drop-george" guy from github:

type User = {
    a: number;
    b: number;
    c: number;
}

type ElementType < T extends ReadonlyArray < unknown > > = T extends ReadonlyArray<
  infer ElementType
>
  ? ElementType
  : never

function get<T, K extends (keyof T)[]>(v: T, keys: K): Pick<T, ElementType<K>> {
    return v
}

const user: User = {a: 1, b: 2, c: 3}
const v = get(user, ['a', 'b'])
const {a, b, c} = v // ERROR
const {a: a2, b: b2} = v // OK
const {a: a3} = v // OK
get({a: 1, b: 2, c: 3} as User, ['a', 'b', 'otherKey']) // ERROR

Original source

UPD: You can just use ElementType<K> instead of ElementType<typeof keys>

UPD2: AFAIK if you want to make something like fun<User>(['firstName', 'lastName']) the only 2 ways are:

• specify properties you need twice:

function get<T, K extends (keyof T)[]>(keys: K): Pick<T, ElementType<K>> {
    // return user
}

const user = get<User, ['firstName', 'lastName']>(['firstName', 'lastName'])

• or make a function that returns another function kinda like this:

function get<T>(): <K extends (keyof T)[]>(keys: K) => Pick<T, ElementType<K>> {
    return keys => {
        // return user from db
    }
}

const user = get<User>()(['firstName', 'lastName'])