打字稿->知道给定的接口是否在类型别名中

Question

I have a type alias on multiple interfaces :

export interface A {}
export interface B {}
export interface C {}
export type Reference = A | B | C;

I have a method called getReference which returns, by default, an array of Reference. I would like my method takes a generic type and check if the given type is in my type alias.

That I have :

export const getReference = (slug: ReferencesSlugs): (state: object) => Array<Reference> => {
   ....... // some code
   // return Reference[]
}

That I want :

Dev can pass generic type and Typescript check if the given type is in the type alias (Reference).

export const getReference = <T>(slug: ReferencesSlugs): (state: object) => Array<T> => {
    ....... // some code
    // As T is in Reference type -> return T[]
}



this.store.pipe( select(getReference<A>('onchonch')) ); // VALID, tslint is ok
this.store.pipe( select(getReference<E>('onchonch')) ); // INAVLID, E is not in my type alias.

Thanks by advance :)

Answer 1

Assuming your union members (the union in the type alias) are not as simple as in your question (ie they have some members) you can use a type constraint on your generic type parameter:

export interface A { a: number; }
export interface B { b: number; }
export interface C { c: number; }
export interface E { e: number; }
export type Reference = A | B | C;

export const getReference = <T extends Reference>(slug: ReferencesSlugs): (state: object) => Array<T> => {
    return null!;
}

getReference<A>('onchonch'); // VALID, tslint is ok
getReference<E>('onchonch'); // INAVLID, E is not in my type alias.

The reason the members matter is because typescript uses structural compatibility to determine type compatibility. This means if two types have the same fields they will be compatible. This also means two empty interfaces are compatible.