Defining at least 1 property of a set in typescript

Question

I see that you can do mutually exclusive properties like this in TS

type A =
  | { a: "common"; m: any; n: undefined }
  | { a: "common"; m: undefined; n: any };

But how do I create a type that has at least 1, but not neither. And without being verbose...

type A = {a: "common"} & ({m: any} | {n: any});

Does that do it? Is there a better way?

score 1 · Answer 1 · answered Sep 03 '20 at 22:24

1

type Combine<A, B> = A | B | A & B
type A = { a: "common" } & Combine<{ m: any }, { n: any }>

answered Sep 03 '20 at 22:24

Rubydesic

3,386
12
27

is this different than `type A = {a: "common"} & ({m: any} | {n: any});` – Thomas Sep 03 '20 at 22:55

Teneff · Accepted Answer · 2020-09-03T23:23:05.077

1

You can create a generic type to require at least one property like this:

type AtLeastOne<T, U extends keyof T = keyof T> = {
    [K in U]-?: Required<Pick<T, K>> & Partial<Omit<T, K>>
}[U]

and then define your A type like this

type A = { a: "common" } & AtLeastOne<{
  m: any;
  n: any;
}>;

but that might be too verbose for you :)

example

edited Sep 03 '20 at 23:23

answered Sep 03 '20 at 23:01

Teneff

30,564
13
72
103

What does `type A = {a: "common"} & ({m: any} | {n: any});` do? – Thomas Sep 03 '20 at 23:11
@Thomas exactly the same as my `type A` (exactly what you intend to) – Teneff Sep 03 '20 at 23:15

Countingstuff · Answer 3 · 2020-09-03T23:39:46.610

That doesn't do it, because for instance { a: "common", m: 1, n: 2 } is assignable to your second A, but not the first, so they're not the same type.

The accepted answer of Does Typescript support mutually exclusive types? defines an XOR type, I think if you do like, maybe replacing the 'never' in the Without type with 'undefined'

type A = { a: "common" } & XOR<{ m: any }, { n: any }>

You have what you want

edit - oh I misread your question.

Defining at least 1 property of a set in typescript

3 Answers3