Make all properties within a Typescript interface optional

If I want to have an explicit AssetDraft interface, I would use a combination of extends and Partial:

interface Asset {
  id: string;
  internal_id: string;
  usage: number;
}

interface AssetDraft extends Partial<Asset> {}

This isn't possible in TypeScript < 2.1 without creating an additional interface with optional properties; however, this is possible by using mapped types in TypeScript 2.1+.

To do this, use the Partial<T> type which TypeScript provides by default.

interface PostDraft {
    asset: Partial<Asset>;
}

Now all the properties on asset are optional, which will allow you to do the following:

const postDraft: PostDraft = {
    asset: {
        id: "some-id"
    }
};

About Partial<T>

Partial<T> is defined as a mapped type that makes every property in the provided type optional (using the ? token).

type Partial<T> = {
    [P in keyof T]?: T[P];
};

Read more about mapped types here and in the handbook.

Deep Partial

If you want a partial implementation that works recursively on objects then you can use the following type in TS 4.1+:

type DeepPartial<T> = {
    [P in keyof T]?: T[P] extends object ? DeepPartial<T[P]> : T[P];
};