How to discriminate union type by checking any object key in the union in TypeScript?-CodePudding

I have a type that looks like this.

type Focus =
    | { sidebar: true }
    | { tabbar: true }
    | { menu: string }

The whole point of this type structure is the convenience of being able to do things like if (focus.sidebar) or if (focus.menu == "open") so I don't want to rewrite my code as if ("sidebar" in focus) and if ("menu" in focus && focus.menu === "open").

This type would technically work for me:

type Focus =
    | { sidebar: true; tabbar?: undefined; menu?: undefined }
    | { tabbar: true; sidebar?: undefined; menu?: undefined }
    | { menu: string; sidebar?: undefined; tabbar?: undefined }

But this is really inconvenient to write.

I've made it this far:

type PartialEmpty<T> = { [A in keyof T]?: undefined }

type UnionToIntersection<U> = (U extends any ? (k: U) => void : never) extends (
    k: infer I
) => void
    ? I
    : never

type Focus2 = PartialEmpty<UnionToIntersection<Focus>>
// {
//     sidebar?: undefined;
//     tabbar?: undefined;
//     menu?: undefined;
// }

But I can't figure out how to intersect and distribute that type over the union... This is pretty close:

type Distribute<T> = T extends any ? Omit<T, keyof Focus2> & Focus2 : never

type Focus3 = Distribute<Focus>

But it doesn't quite work...

const focus: Focus3 = {} as any
if (focus.menu) {
    focus.m
}

Playground Link

CodePudding user response：

T in Distribute<T> is one of:

| { sidebar: true }
| { tabbar: true }
| { menu: string }

So you need to omit keyof T from Focus2 and intersect with T:

type Distribute<T> = T extends any ? Omit<Focus2, keyof T> & T : never

Playground