Using discriminated unions of objects with union types in TypeScript

Looking for advice on how to create a specific type definition based on species.

Take a look at my current type definition:

type Animal =
  | {
    'species': 'cat'
    'action': 'meow'
  }
  | {
    'species': 'cat'
    'action': 'whine'
  }
  | {
    'species': 'dog' | 'tree'
    'action': 'bark'
  };

I need help defining a conditional type ActionsFor<S> that will narrow down the type based on the given species. For example:

type CatActions = ActionsFor<'cat'> // should be 'meow' | 'whine'
type DogActions = ActionsFor<'dog'> // should be 'bark'
type TreeActions = ActionsFor<'tree'> // should be 'bark'
type BarkActions = ActionsFor<'dog'|'tree'> // should be 'bark'

My current implementation is almost there, but does not handle unioned species correctly:

type ActionFor<S extends Animal['species']> = Extract<Animal, {species: S}>['action']

This leads to:

type CatActions = ActionsFor<'cat'> // correct - 'meow' | 'whine'
type DogActions = ActionsFor<'dog'> // WRONG - never
type TreeActions = ActionsFor<'tree'> // WRONG - never
type BarkActions = ActionsFor<'dog'|'tree'> // correct - 'bark'

Any suggestions on how to redefine ActionsFor to achieve the desired outcome?

Answer №1

Just cracked the code on this one. It appears to be functioning as expected, but I'm open to any suggestions for streamlining or making it more refined!

type ActionFor<S extends Animal['species']> = Exclude<Animal, {species: Exclude<Animal['species'], S>}>['action']

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