如何使用 RxJS 在 Angular 6 中发出一系列 http 请求

Question

我一直在网上寻找解决方案，但找不到任何适合我的用户案例的解决方案。我正在使用 MEAN 堆栈（Angular 6）并且我有一个注册表单。我正在寻找一种方法来执行对 API 的多个 HTTP 调用，每个调用都依赖于前一个返回的结果。我需要看起来像这样的东西：

firstPOSTCallToAPI('url', data).pipe(
    result1 => secondPOSTCallToAPI('url', result1)
    result2 => thirdPOSTCallToAPI('url', result2)
    result3 => fourthPOSTCallToAPI('url', result3)
    ....
).subscribe(
    success => { /* display success msg */ },
    errorData => { /* display error msg */ }
);

我需要使用哪种 RxJS 运算符组合来实现这一点？一种可能的解决方案是嵌套多个订阅，但我想避免这种情况并使用 RxJS 做得更好。还需要考虑错误处理。

Answer 1

对于依赖于先前结果的调用，您应该使用concatMap

firstPOSTCallToAPI('url', data).pipe(
    concatMap(result1 => secondPOSTCallToAPI('url', result1))
      concatMap( result2 => thirdPOSTCallToAPI('url', result2))
       concatMap(result3 => fourthPOSTCallToAPI('url', result3))
    ....
).subscribe(
    success => { /* display success msg */ },
    errorData => { /* display error msg */ }
);

如果你的异步方法不依赖于前面异步调用的返回值，你可以使用

   concat(method(),method2(),method3()).subscribe(console.log)

Answer 2

我遇到了同样的问题，这是我的解决方案，使用 pipe 和 concatMap 作为数组获取开始时间和结束时间之间的时间段的序列数据。

这是我们有数组请求时的通用解决方案。

我为谁担心。

 let currentReplayData = [];
 let timerange = [[t1, t2], [t3, t4]]; // array of start and end time
 from(timerange).pipe(
      concatMap(time => <Observable<any>>this.dataService.getData(time[0],time[1]))
      ).subscribe(val => {
        //console.log(val)
        this.currentReplayData = this.currentReplayData.concat(val);
      });

Answer 3

MergeMap

正是你要找的东西

firstPOSTCallToAPI('url', data).pipe(
    mergeMap(result1 => secondPOSTCallToAPI('url', result1)),
    mergeMap(result2 => thirdPOSTCallToAPI('url', result2)),
    mergeMap(result3 => fourthPOSTCallToAPI('url', result3)),
    // ...
).subscribe(
    success => { 
      // here you will get response of LAST request (fourthPOSTCallToAPI)
    },
    errorData => { /* display error msg */ }
);


// I assume that
// secondPOSTCallToAPI, thirdPOSTCallToAPI and fourthPOSTCallToAPI
// returns obserwable eg. return this.http.get<Book>(apiUrl);

Answer 4

import { HttpClient } from '@angular/common/http';
import { Injectable } from '@angular/core';
import { Observable } from 'rxjs/Observable';
import 'rxjs/add/observable/forkJoin';

@Injectable()
export class DataService {

  constructor(private http: HttpClient) { }

  public requestDataFromMultipleSources(): Observable<any[]> {
    let response1 = this.http.get(requestUrl1);
    let response2 = this.http.get(requestUrl2);
    let response3 = this.http.get(requestUrl3);
    return Observable.forkJoin([response1, response2, response3]);
  }
}

The above example shows making three http calls, but in a similar way you can request as many http calls as required

    import { Component, OnInit } from '@angular/core';
import { DataService } from "../data.service";

@Component({
    selector: 'app-page',
    templateUrl: './page.component.html',
    styleUrls: ['./page.component.css']
})
export class DemoComponent implements OnInit {
    public responseData1: any;
    public responseData2: any;
    public responseData3: any;

    constructor(private dataService: DataService) {}

    ngOnInit() {
        this.dataService.requestDataFromMultipleSources().subscribe(responseList => {
            this.responseData1 = responseList[0];
            this.responseData2 = responseList[1];
            this.responseData3 = responseList[1];
        });
    }
}

Answer 5

让我在这里向您展示如何操作，假设我有很多要按顺序发送电子邮件的电子邮件：

sendEmails() {
  this.isLoading = true; 
            const calls = this.emails <<-- assume this contain an array of emails
            .map(email => this.userEmailService.deliver({email: email, userId: 1242}));
            from(calls) <<-- make use of the from.
                .pipe(
                    concatMap(res => res),
                    finalize(() => this.isLoading = false)
                ).subscribe(() => { });
}

我希望这会有所帮助。

Answer 6

试试这个，Angular 提供了一次调用多个 API 的功能。

forkJoin()

您将在array 中以与调用 API 相同的顺序获取数据。

例如：

forkJoin(request1, request2)
    .subscribe(([response1, response2]) => {

你可以找到更多read

我还给了另一个answer。请检查一下，它也可能对您有所帮助。