How can one generate a mapped, conditional type in TypeScript that eliminates properties of type string | null?

Here's my latest unsuccessful attempt before seeking guidance. Can anyone point out my mistakes? Thank you!

interface Entity {
  id: string;
  title: string | null;
}

interface AnotherEntity {
  id: string;
  title: string;
}

type ExcludeNullFields<T> = {
  [K in keyof T as T[K] extends string | null ? K : never]: T[K];
};

type TitlelessEntity = ExcludeNullFields<Entity>; 
// expected result: { id: string }

type TitledEntity = ExcludeNullFields<AnotherEntity>; 
// expected result: { id: string; title: string }

Answer №1

When you modify ExcludeNullableFields like this...

type ExcludeNullableFields<A> = {
  [K in keyof A as Extract<A[K], null> extends  never ? K : never]: A[K] ;
};

The main idea here is to determine whether null can be extracted from any of the types. If a type does not include null, the Extract function will return never. However, if Extract does find the type null, then it indicates that the field is nullable.

Answer №2

Your ExcludeNullableFields conditional type seems to be including fields that are of type string | null, when it should actually be excluding them. To fix this, you'll need to adjust the logic within your conditional:

type ExcludeNullableFields<T> = {
  [Key in keyof T as T[Key] extends string ? Key : never]: T[Key];
};

type UnnamedItem = ExcludeNullableFields<SomeItem>; 
// output: { id: string }

type NamedItem = ExcludeNullableFields<AnotherItem>; 
// output: { id: string; name: string }

I hope this explanation clarifies things for you.

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