CODEWITHSUNDEEP
typescript
OliverRadini
OliverRadini

Reputation: 6466

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

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?

Upvotes: 1

Views: 101

Answers (1)

Titian Cernicova-Dragomir
Titian Cernicova-Dragomir

Reputation: 250106

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

Upvotes: 1

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