Understanding the return type of a function in TypeScript

Question

Understanding the return type of a function in TypeScript

As someone new to TypeScript, I have some knowledge about the concept of generics in Java. Here's my dilemma: I have three functions - searchTrack, searchAlbum, and searchArtist.

  searchTrack(query: string): Observable<Track[]> {
    return this.search(query, 'track');
  }

  searchArtist(query: string): Observable<Artist[]> {
    return this.search(query, 'artist');
  }

  searchAlbum(query: string): Observable<Album[]> {
    return this.search(query, 'album');
  }

I am looking for a general function named 'search' within this class that can take the query and the type of entity as parameters, and then return an observable collection of a specific entity type. However, I am unsure how to utilize generics in order to specify a generic return type for a function.

search(query: string, type: string): Observable<Array<T>> {
 return this.query(`/search`, [
   `q=${query}`,
   `type=${type}`
 ]);
}

I'm seeking guidance on whether there is a way to achieve this functionality?

function generics typescript

Answer 1

Answer №1

Consider using the Array class instead of []. It is also recommended to define a generic T type on the search function.

search<T>(query: string, type: string): Observable<Array<T>> {
 return this.query(`/search`, [
   `q=${query}`,
   `type=${type}`
 ]);
}

Ensure you are calling it correctly like this:

let result = search<Artist>('someQuery', 'artist');

To learn more about generics in typescript, check out the Generics chapter in the handbook here.

Answer 2

Consider using the Array class instead of []. It is also recommended to define a generic T type on the search function.

search<T>(query: string, type: string): Observable<Array<T>> {
 return this.query(`/search`, [
   `q=${query}`,
   `type=${type}`
 ]);
}

Ensure you are calling it correctly like this:

let result = search<Artist>('someQuery', 'artist');

To learn more about generics in typescript, check out the Generics chapter in the handbook here.

Answer 3

Answer №2

When @toskv provided an answer, he mentioned that you can include a generics type in the method signature. However, the compiler cannot automatically infer the type, so you must specify it like this:

myObj.search<Track>(query, "track");

Alternatively, you can implement something similar to the following:

interface MyObservableClass {}

interface MyObservableClassCtor<T extends MyObservableClass> {
    new (): T;
    getType(): string;
}

class Artist implements MyObservableClass {
    static getType(): string {
        return "artist";
    }
}

class Album implements MyObservableClass {
    static getType(): string {
        return "album";
    }
}

class Track implements MyObservableClass {
    static getType(): string {
        return "track";
    }
}

class Searcher {
    search<T extends MyObservableClass>(ctor: MyObservableClassCtor<T>, query: string): Observable<T[]> {
        return this.query(`/search`, [
            `q=${query}`,
            `type=${ ctor.getType() }`
        ]);
    }
}

let searcher: Searcher = ...;

searcher.search(Track, "...");

This way, the compiler can deduce the type T by providing it with the corresponding class (/ctor).

Answer 4

When @toskv provided an answer, he mentioned that you can include a generics type in the method signature. However, the compiler cannot automatically infer the type, so you must specify it like this:

myObj.search<Track>(query, "track");

Alternatively, you can implement something similar to the following:

interface MyObservableClass {}

interface MyObservableClassCtor<T extends MyObservableClass> {
    new (): T;
    getType(): string;
}

class Artist implements MyObservableClass {
    static getType(): string {
        return "artist";
    }
}

class Album implements MyObservableClass {
    static getType(): string {
        return "album";
    }
}

class Track implements MyObservableClass {
    static getType(): string {
        return "track";
    }
}

class Searcher {
    search<T extends MyObservableClass>(ctor: MyObservableClassCtor<T>, query: string): Observable<T[]> {
        return this.query(`/search`, [
            `q=${query}`,
            `type=${ ctor.getType() }`
        ]);
    }
}

let searcher: Searcher = ...;

searcher.search(Track, "...");

This way, the compiler can deduce the type T by providing it with the corresponding class (/ctor).

Answer 5

Answer №3

If you want to handle different types of search queries, you can use overloads in your code:

class Search {
    find<T>(query: string, type: "Movie"): Observable<Movie[]>;
    find<T>(query: string, type: "Book"): Observable<Book[]>;
    find<T>(query: string, type: "Song"): Observable<Song[]>;
    find<T>(query: string, type: string): Observable<T[]>;
    find<T>(query: string, type: string): Observable<T[]> {
        return null;
    }
}

This setup ensures that the correct type is returned based on the query provided:

var result = searcher.find("Harry Potter", "Book");

Trying to assign incompatible objects will trigger a compilation error:

var errorResult: Observable<Song[]> = searcher.find("Titanic", "Book");

Answer 6

If you want to handle different types of search queries, you can use overloads in your code:

class Search {
    find<T>(query: string, type: "Movie"): Observable<Movie[]>;
    find<T>(query: string, type: "Book"): Observable<Book[]>;
    find<T>(query: string, type: "Song"): Observable<Song[]>;
    find<T>(query: string, type: string): Observable<T[]>;
    find<T>(query: string, type: string): Observable<T[]> {
        return null;
    }
}

This setup ensures that the correct type is returned based on the query provided:

var result = searcher.find("Harry Potter", "Book");

Trying to assign incompatible objects will trigger a compilation error:

var errorResult: Observable<Song[]> = searcher.find("Titanic", "Book");

Understanding the return type of a function in TypeScript

Answer №1

Answer №2

Answer №3

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