CODEWITHSUNDEEP
angulartypescriptrxjs
Kris
Kris

Reputation: 912

angular2 map data as specific object type

I created a very simple app based on the Angular2 tutorial.

To start, I have a very simple "Book" model:

 /**
 * book model
 */
export class Book {
    public data;

    /**
     * constructor
     * @param id
     * @param title
     * @param pages
     */
    constructor(
        public id,
        public title:string,
        public pages:Array
    ){
        alert('it works'); // just a check
    }
}

In my service, I get a book like this:

return this._http.get('getBook/1')
        .map(function(res){
            return <Book> res.json();
        })

My expectation was that this would get the resulting JSON data and "map" it to a Book object.

However, it just returns an object with type "Object".

I could create a new Book object myself and pass the parameters in the constructor, like this:

return new Book(res.id, res.title, res.pages);

Is this the best way to do this? Did I miss something?

Upvotes: 11

Views: 52659

Answers (4)

Tenzin Chemi
Tenzin Chemi

Reputation: 5871

From https://angular.io/guide/http#requesting-a-typed-response

To specify the response object type, first define an interface with the required properties. Use an interface rather than a class, because the response is a plain object that cannot be automatically converted to an instance of a class.

Always receive response using interface because only thing that's happening is mapping fields, and you cannot use class if you have instance function which will throw error when you use that object.

product.getPrice is not a function

Example

interface ProductInterface {
  id: number;
  name: string;
  price: number;
}

class Product implements ProductInterface {
  id: number;
  name: string;
  price: number;
  
  constructor(productData: ProductInterface) {
    this.id = product.id;
    this.name = product.name;
    this.price = product.price;
  }
  
  public getPrice(): string {
    return this.price + " INR";
  }
}

class ProducService {
  ...
  ...


  getProduct(): Observable<Product> {
    return this.http.get<Product>('assets/product.json').pipe(map(data => new Product(data))); 
  }

  getProductWithoutMappingToClass(): Observable<Product> {
    return this.http.get<Product>('assets/product.json');
  } // Throw runtimerror ctx_r1.product.getPrice is not a function
}

Upvotes: 2

Przemek Struciński
Przemek Struciński

Reputation: 5238

Good practice is to consume data from GET response using 

Observable<Model>

(regarding to Angular documentation https://angular.io/guide/http) And then:

// imports

import {HttpClient} from "@angular/common/http";

// in constructor parameter list

private http: HttpClient

// service method

getBook(): Observable<Book> {return this.http.get<Book>({url}, {options});}

Upvotes: 5

dieuhd
dieuhd

Reputation: 1326

I think you should declare an interface Book instead of class book:

export interface Book {
    public id;
    public title:string;
    public pages:Array;
}

In your service:

//get one record
return this._http.get('getBook/1')
        .map(function(res){
            return <Book> res.json();
        });

//get multi record
return this._http.get('getBooks')
        .map(function(res){
            return <Book> res.json();
        });

Upvotes: 5

Thierry Templier
Thierry Templier

Reputation: 202216

Yes, casting an object to a type in TypeScript doesn't create an instance of this type. It's just a facility of TypeScript for type checking.

If you want actually an instance of Book you need to use something like that:

return this._http.get('getBook/1')
    .map(function(res){
        var data = res.json();
        return new Book(data.id, data.title, data.pages);
    })

To answer your question. In fact if you only have fields into your type (with an interface for example), casting is enough. Moreover if you have methods you want to use later, it's necessary to implicitly create an instance of the Book type (see above) instead of casting. Otherwise you won't be able to use them (they will be undefined on your object)...

See this question for more details:

Upvotes: 15

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