角4- HttpClient无法隐藏状态代码错误消息

Question

我把请求发送给服务器。我为请求添加了错误处理程序，但是我在下面得到了控制台错误：

发布http://localhost:7777/auth 401 (未经授权)

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我发送401代码从服务器进行测试，现在我不知道有删除这条消息。

import {HttpClient} from "@angular/common/http";
import {WebStorage} from "../singleton/web-storage";
import {environment} from "../../environments/environment";
import * as $ from "jquery/dist/jquery.slim";
import {Injectable} from "@angular/core";
import {Observable} from "rxjs/Observable";
import * as jwtDecode from 'jwt-decode';
import {RequestHandler} from "../handler/request.handler";

@Injectable()
export class Auth {

    constructor(private httpClient: HttpClient,
                private requestHandler: RequestHandler) {

    }

    /**
     * Check is user authorized
     *
     * @returns {Observable<Response>}
     */
    isAuthorized(): Observable<Response> {
        return new Observable((observer) => {
            let formData: Object = $.param({
                transferData: {
                    jwt: WebStorage.ssRetrieve('jwt')
                }
            });

            this.httpClient.post<Object>(environment.getUrl('/auth'), formData, {
                headers: {
                    'Content-Type': 'application/x-www-form-urlencoded; charset=UTF-8'
                }
            })
                .subscribe(
                    (response: Response) => {
                        observer.next(response);
                    },
                    error => {}// For this example, error handler is empty, just for hide most console errors
                );
        });
    }
}

任何建议！

Answer 1

您可以将可观察到的捕捉属性用作：

this.httpClient.post(your_url).catch((ex,caught) => {
 console.log('An exception has been caught!');
 return Observable.empty();
})

在这里，您可以抛出错误，而不是返回空的可观察到的，但这正是您希望避免的。因此，返回空可观察解决了你的情况。