在 Angular2 中捕获 401 异常

Question

当我尝试连接到未经授权的 URL 时，我进入 Chrome:

zone.js:1274 POST http://localhost:8080/rest/v1/runs 401 (Unauthorized)
core.umd.js:3462 EXCEPTION: Response with status: 401 Unauthorized for URL: http://localhost:8080/rest/v1/runs

我的Home组件的代码是：

import {Component, OnInit} from '@angular/core';
import {Run} from "../_models/run";
import {Http, Response} from "@angular/http";
import {RunService} from "../_services/run.service";
import {Observable} from "rxjs";

@Component({
    moduleId: module.id,
    templateUrl: 'home.component.html'
})

export class HomeComponent implements OnInit{
    url: "http://localhost:8080/rest/v1/runs"
    username: string;
    runs: Run[];

    constructor(private http: Http, private runService: RunService) {

    }

    ngOnInit(): void {
        this.username = JSON.parse(localStorage.getItem("currentUser")).username;
        this.runService.getRuns()
            .subscribe(runs => {
                this.runs = runs;
            });
    }
}

并且此组件使用此服务：

import { Injectable } from '@angular/core';
import {Http, Headers, Response, RequestOptions, URLSearchParams} from '@angular/http';
import { Observable } from 'rxjs';
import 'rxjs/add/operator/map'
import {AuthenticationService} from "./authentication.service";
import {Run} from "../_models/run";

@Injectable()
export class RunService {
    url = "http://localhost:8080/rest/v1/runs";
    private token: string;

    constructor(private http: Http, private authenticationService: AuthenticationService) {

    }

    getRuns(): Observable<Run[]> {
        return this.http.post(this.url, JSON.stringify({ token: this.authenticationService.token }))
            .map((response: Response) => {
                console.log(response.status);
                if (response.status == 401) {
                    console.log("NOT AUTHORIZED");
                }

                let runs = response.json();
                console.log(runs);
                return runs;
            });
    }
}

捕获此 401 异常的正确方法是什么？我应该在哪里执行此操作？在组件中还是在服务中？最终目标是在发生任何 401 响应时重定向到登录页面。

Answer 1

您很可能希望从您的 RunService 中抛出一个错误，该错误可以在您的组件中捕获，该组件可以路由到登录页面。下面的代码应该可以帮助您：

在运行服务中：

需要从 rxjs 导入 catch 运算符：

import 'rxjs/add/operator/catch';

并且您的 getRuns() 函数应更改为

getRuns(): Observable<Run[]> {
    return this.http.post(this.url, JSON.stringify({ token: this.authenticationService.token }))
        .map((response: Response) => {
            let runs = response.json();
            return runs;
        })
        .catch(e => {
            if (e.status === 401) {
                return Observable.throw('Unauthorized');
            }
            // do any other checking for statuses here
        });

然后组件中的 ngOnInit 将是：

ngOnInit(): void {
    this.username = JSON.parse(localStorage.getItem("currentUser")).username;
    this.runService.getRuns()
        .subscribe(runs => {
            this.runs = runs;
        }, (err) => {
            if (err === 'Unauthorized') { this.router.navigateByUrl('/login');
        });
}

显然，您需要根据自己的需要调整代码并根据需要更改它，但是从 Http 捕获错误、抛出 Observable 错误并使用 err 回调来处理您的错误的过程组件应该可以解决您的问题。

在 Angular2 中捕获 401 异常

Catch 401 Exception in Angular2

exception

http-status-code-401

angular