Conditional type based on one property

Question

When actions is defined in the object, I want type C to be B, when it's not defined, I want type C to be A, how to implement such type/interface?

interface A {
  label: string
  prop: string
}

interface B {
  label?: string
  actions: string[]
}

// Pseudocode
type C = actions ? B : A

const c1: C = { //  type A
  label: 'c1',
  prop: 'c1',
}

const c2: C = { //  type B
  label: 'c2',
  actions: ['c2'],
}

Answer 1

You'll need a type parameter on C so that it knows what type its dealing with, which for the assignments shown makes C fairly pointless (just use A or B). You also can't infer type arguments in simple assignments like those.

You can with functions, though, so if you're really writing a function and just simplified it for the question (or you're happy to have a pass-through function), you can do this:

type C<T> = T extends { actions: string[] } ? B : A;

function example<T>(data: T): C<T> {
    return data as C<T>;
}

const c1 = example({
    label: "c1",
    prop: "c1",
});
console.log(c1);
//          ^? const c1: A

const c2 = example({
    label: "c2",
    actions: ["c2"],
});
console.log(c2);
//          ^? const c2: B

Playground link