Type that omits all inherited properties?

Say we have

class Person {
  age: number
}
class Child extends Person {
  play() {
    //...
  }
}

class Person {
  age: number
}
class Child extends Person {
  play() {
    //...
  }
}

I want a type that incudes only the fields from Child without the inherited properties. For example, type OnlyChild = { play: () => void }. Is that possible. I know I can use Omit<> but I don't know what the properties of Person are.

3 Replies

bentenOP•3y ago

wait type OnlyNonInherited = Omit<B, keyof A>; maybe? Except I don't know A

Unknown User•3y ago

Message Not Public

Froxx•3y ago

If you think about types like types instead of interfaces they will extend each other by doing type B = A & {bProp: any}, so in that syntax compared to the actual extend one used in interfaces, you can see it's pretty much just a concatenation of multiple types, not a layering. Therefore I think this might not be possible. May I ask you for the purpose of that? There might be a different approach doing what you're aiming for

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Type that omits all inherited properties?