TypeScript Exclude an Extended Interface

Question

I have an interface which acts as a parent of multiple child interfaces. For example:

interface ParentProps {
  family?: ChildA & ChildB
}

What I would like to do, ideally, is in certain situations exclude either ChildA or ChildB interfaces from the family prop.

I have tried some ideas but with errors.

interface FamTree extends Exclude<ParentProps['family'], ChildA> {}

An interface can only extend an object type or intersection of object types with statically known members

When I try doing this as a type instead of an interface, I don't get any errors, but I also don't get the correct types, either.

type FamTree = Exclude<ParentProps['family'], ChildA>;

This results in FamTree having a type of never though. Not sure how best to approach this.

The end result I am looking to achieve is:

type FamTree = {
  family?: ChildB
}

Answer 1

To solve this, you can Exclude undefined from ParentProps['family'] and afterwards Omit all keys of ChildA from the resulting type.

type FamTree = Omit<Exclude<ParentProps['family'], undefined>, keyof ChildA>

Playground

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If you want to keep shared properties, you have to Exclude all the keys from A which B also has.

type FamTree = Omit<
  Exclude<ParentProps['family'], undefined>, Exclude<keyof ChildA, keyof ChildB>
>

Playground