Angular 4 Http Interceptor:next.handle(...).do不是一个函数

Question

我创建了这个HTTPInterceptor以便能够更好地处理http错误,它在我执行git pull并运行npm install之前运行良好.

这是我的代码:

import {Injectable} from '@angular/core';
import {HttpEvent, HttpInterceptor, HttpHandler, HttpRequest, HttpResponse} from '@angular/common/http';
import {Observable} from "rxjs";
import {ToasterService} from "angular2-toaster";

@Injectable()
export class GobaeInterceptor implements HttpInterceptor {
    constructor(private toasterService: ToasterService){
    }
    intercept(req: HttpRequest<any>, next: HttpHandler): Observable<HttpEvent<any>> {
        return next.handle(req)
            .do(event => {
                if (event instanceof HttpResponse) {
                    let response = event.body;
                    if(response.Error){
                        this.toasterService.pop('error', 'Error '+response.Code, response.Message);
                    }
                }
            });
    }
}

这是我得到的错误:

TypeError:next.handle(...).do不是GobaeInterceptor.webpackJsonp中的函数.../../../../../src/app/services/gobae.interceptor.ts.GobaeInterceptor.拦截(gobae.interceptor.ts:12)在HttpInterceptorHandler.webpackJsonp .../../../common/@angular/common/http.es5.js.HttpInterceptorHandler.handle(

最近有什么可以影响我的代码的东西改变了吗？我现在可以做什么来"捕获"拦截器上的http响应？

Answer 1

抛出此错误是因为您缺少do运算符.下面导入使用do运算符修补可观察对象.

import 'rxjs/add/operator/do';

默认情况下,RxJs不会捆绑所有运算符函数以减小库大小.您需要导入要单独使用的运算符.

Answer 2

rxjs 6/angular 6将需要管道

return next.handle(req).pipe(
  tap(event => {
    if (event instanceof HttpResponse) {
      ...
    }
  })
);

Answer 3

你必须使用导入.

import 'rxjs/add/operator/do';
import 'rxjs/add/operator/catch';
import 'rxjs/Observable';
import 'rxjs/add/observable/throw';