I have this simple Component
import {Component} from 'angular2/core';
import {RouterLink, RouteParams} from 'angular2/router';
import {Http, Response, Headers} from 'angular2/http';
import {User} from '../../models/user';
import 'rxjs/add/operator/map';
@Component({
template: `
<h1>{{user.name}} {{user.surname}}</h1>
`,
directives: [RouterLink],
})
export class UserComponent {
user: User;
constructor(private routeParams: RouteParams,
public http: Http) {
this.user = new User();
this.http.get('http://localhost:3000/user/' + this.routeParams.get('id'))
.map((res: Response) => res.json())
.subscribe((user: User) => this.user = user);
console.log(this.user);
}
}
Why does subscribe
not cast the response into a full User
object. When I am logging the user
variable my console say User {_id: undefined, name: undefined, surname: undefined, email: undefined}
. But nevertheless binding to .name
and .surname
in the view is working..
What happens here? Where is the user actually stored?
Found a solution here: https://stackoverflow.com/a/29759472/2854890
My Method now looks like this:
constructor(private routeParams: RouteParams,
public http: Http) {
this.user = new User();
this.http.get('http://localhost:3000/user/' + this.routeParams.get('id'))
.map((res: Response) => res.json())
.subscribe((json: Object) => {
this.user = new User().fromJSON(json);
});
}
I enhanced the Serializable
by returning the object in the end, so I can leave out something like
var u = new User();
u.fromJSON(...);
and just write
new User().fromJSON(json);
Serializable
class
export class Serializable {
fromJSON(json) {
for (var propName in json)
this[propName] = json[propName];
return this;
}
}
Good practice is to consume data from GET response using
Observable<Model>
(regarding to Angular documentation https://angular.io/guide/http)
so...
// imports
import {HttpClient} from "@angular/common/http";
// in constructor parameter list
private http: HttpClient
// service method
getUser(): Observable<User> {return this.http.get<User>({url}, {options});}
You do not need to do anything more. I consider this approach as most friendly.
That's not supported by TypeScript.
See How do I cast a JSON object to a typescript class for more details.