Make all properties within a Typescript interface optional

I have an interface in my application:

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

that is part of a post interface:

interface Post {
  asset: Asset;
}

I also have an interface that is for a post draft, where the asset object might only be partially constructed

interface PostDraft {
  asset: Asset;
}

I want to allow a PostDraft object to have a partial asset object while still checking types on the properties that are there (so I don't want to just swap it out with any).

I basically want a way to be able to generate the following:

interface AssetDraft {
  id?: string;
  internal_id?: string;
  usage?: number;
}

without entirely re-defining the Asset interface. Is there a way to do this? If not, what would the smart way to arrange my types in this situation be?

This isn't possible in TypeScript type which TypeScript provides by default.

interface PostDraft {
    asset: Partial;
}

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

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

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

Stack Overflow Q#39713349, score: 465