How to define an type such that it only needs to implement one property of another interface

Question

If I have an interface:

interface ITest {
    a: number;
    b: string;
}

I want to implement an interface such that a function needs to take in an object with only some defined subset of the properties of the interface, for instance for the above example, it may only need to take in a . I've tried the following:

type WithOnly<T, K extends keyof T> = {
    [K in keyof T]: T[K];
}

export const f = (x: WithOnly<ITest, 'a'>) => settings.a * 2;

But the compiler doesn't seem to like this; it wants x to have b also. Is there a way to implement this?

Answer 1

You are close, since you only want the properties specified in K you should do [P in K]: T[P];

type WithOnly<T, K extends keyof T> = {
    [P in K]: T[P];
}

Play

Or better yet just use Pick<ITest, 'a'> which already does what you want it to do:

interface ITest {
    a: number;
    b: string;
}
export const f = (x: Pick<ITest, 'a'>) => x.a * 2;

f({ a: 1 })

Play