Extract property type from a oneof

I’ve got a type a such:

type OneOf =
  | {
      $case: 'one';
      one: One;
    }
  | {
      $case: 'two';
      two: Two;
    }
  | {
      $case: 'three';
      three: Three;
    };

I want to create a new type to be able to get from the input OneOf, get the output One | Two | Three.

I have tried many different things, including one from this answer with UnionToIntersection in the hope that with an intersection, I could simply exclude the $case, loop on the keys and return the type.

I always get a never type as output though.

Essentially, I thought I could get away with it with the following:

type ExtractOneOfTypes<T, K extends keyof T = keyof T> = {
  [key in K]: T extends { $case: key } ? T[key] : never;
}[K];

Because in my oneOf, for each $case, on that same object there is a property name that is the same as the $case value. But this doesn’t work. Any ideas?

>Solution :

You you can infer the literal value type of $case and then check if it’s a keyof T using a nested Conditional Type like so:

type One = { one: string };
type Two = { two: number };
type Three = { three: boolean };

type OneOf =
  | { $case: "one"; one: One }
  | { $case: "two"; two: Two }
  | { $case: "three"; three: Three };

type ExtractOneOfTypes<T, K extends keyof T> = T extends { [Key in K]: infer U }
  ? U extends keyof T
    ? T[U]
    : never
  : never;

type ExtractedType = ExtractOneOfTypes<OneOf, "$case">;
//    ^? type ExtractedType = One | Two | Three

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