Typescript - How to cast object to interface

Question

Suppose there is an interface

interface A {
  a: number;
}

and currently I have a raw object

const a = { a: 1, b: '2' };

I expect a way to get x to be { a: 1 }, which contains only members defined in interface.

This can be achieved through creating an empty {} and looping over and filling in all A's members, but considering interface members and object's redundant fields can be up to dozens, it is not practical to do so.

I tried

let x = <A> a; // or let x = a as A;
console.log((x as any).b); // still exist

I also looked up in lodash and found keys and pick, but these are targeted for class instance instead of interface.

Answer 1

Interfaces don't exist at runtime. Your best option would be delete unwanted properties yourself or create a copy of the object with the properties you need.

For example you can use a method like this (method copied from this answer):

function pick<T, K extends keyof T>(obj: T, ...keys: K[]): Pick<T, K> {
    const copy = {} as Pick<T, K>;

    keys.forEach(key => copy[key] = obj[key]);

    return copy;
}

const a = { a: 1, b: '2' };
let copy : A = pick(a, "a");
console.log(copy);

Answer 2

Basically your question boils down to the problem of getting all properties of the interface as an array of strings and then make an copy of the object containing only those properties. Making a copy of object is pretty simple and there are tons of examples out there. But the first issue is much harder to tackle.

As far as I know the only way to get this done "now" is to use custom transformers for typescript during compilation. Here is an example, but please note - this require to use typescript 2.4+ that is not yet in stable release.