Typescript Pick<> using types rather than keys

Question

I under stand how the pick type works in typescript. How ever I want a version of it that will pick all the strings out of an interface rather than using the keys.

interface Nested {
    bla: string;
    other: number;
}

interface Testing {
    bla: string;
    other: number;
    list: Array<string>;
    list2: Array<string>;
    list3: Array<any>;
    list4: Array<number>;
    objecter: Nested
    objecter2: object
}


type NewPickType<T,Type> = // Some Typescript interface Logic 

type StringArrayOnlyType = NewPickType<Testing,Array<string>>

let obj: StringArrayOnlyType = {
    bla: "", // error
    other: 0, // error
    list: [""], // ok 
    list2: [""], // ok
    list3: [""], // ok
    list4: [0], // error
    objecter: { bla: "", other: 0 }, // error
    objecter2: {} // error
}

Any ideas on how to build the NewPickType

mullin · Accepted Answer · 2022-04-09 09:44:48Z

1

Ended up figuring it out just after posting but this might help some other people.

export type PickTypeKeys<Obj, Type, T extends keyof Obj = keyof Obj> = ({ [P in keyof Obj]: Obj[P] extends Type ? P : never })[T];

export type PickType<T, Type> = Pick<T, PickTypeKeys<T, Type>>
export type OmitType<T, Type> = Omit<T, PickTypeKeys<T, Type>>

Needed to use two steps on to get they keys of the type I was looking for and one to filter the object using those keys.

Example

edited Apr 9, 2022 at 9:44

answered Jun 22, 2021 at 3:04

mullin

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