Why does finally() not work after catching on an Observable in Angular 2? : <p>I am trying to add a loading spinner to every request that ends in Angular 2, so I extended the Http service calling it HttpService. After every request I would like to call a finally() function after catching errors so that I can stop the loading spinner.</p>

<p>But typescript says: </p>

<blockquote>
  <p>[ts] Property 'finally' does not exist on type 'Observable'.</p>
</blockquote>

<pre><code>import { AuthService } from './../../auth/auth.service';
import { Injectable } from '@angular/core';
import { Http, XHRBackend, RequestOptions, RequestOptionsArgs, Request, Response, Headers } from '@angular/http';

import { Observable } from 'rxjs/Observable';
import 'rxjs/add/operator/catch';
import 'rxjs/add/observable/throw';

@Injectable()
export class HttpService extends Http {

    constructor(
        backend: XHRBackend,
        options: RequestOptions,
        private authservice: AuthService
    ) {
        super(backend, options);
        this.updateHeaders(options.headers);
    }

    request(url: string|Request, options?: RequestOptionsArgs): Observable&lt;Response&gt; {
        return super.request(url, options)
            .catch((response: Response) =&gt; this.authError(response))
            .finally(() =&gt; {
                // ...
                /* Do something here when request is done. For example
                finish a spinning loader. */
            });
    }

    private authError(response: Response) {
        if (response.status === 401 || response.status === 403) {
            this.authservice.logout();
        }
        return Observable.throw(response);
    }

    private updateHeaders(headers: Headers) {
        headers.set('Content-Type', 'application/json');
        if (this.authservice.isloggedIn()) {
            headers.set('Authorization', `Bearer ${this.authservice.getToken()}`);
        }        
    }
}
</code></pre>

<p>How can I run some code after every http request like this? What is the best way of doing it?</p>
