尝试比较"..."时出错。只允许数组和可迭代对象。(使用 switchMap)



所以我正在研究Angular,我正在使用Spotify API进行一个项目。当我搜索音乐时,我收到此错误(尝试差异"A$AP Twelvyy"时出错。只允许数组和可迭代对象(。我想使用 switchMap,因为事件通过 keyup 触发。

这是服务


export class SpotifyService{
    constructor(private _http: HttpClient){
    }
    searchMusic(query: string){
      debugger;
      const searchUrl=`https://api.spotify.com/v1/${query}`;

      const headers=new HttpHeaders({
        Authorization:
        "Bearer XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXx"
      });
      return this._http.get(searchUrl, {headers});
    }

          getArtists(query: string) {
            debugger
            return this.searchMusic(`search?q=${query}&type=artist&limit=15`).pipe(
              switchMap(data => data["artists"].items)
            );
          }
}

这是搜索组件

import { Component } from '@angular/core';
import {SpotifyService} from '../services/spotify.services'

@Component({
    selector: 'app-searchbar',
    templateUrl: './searchbar.component.html',
    styleUrls: ['./searchbar.component.scss'],
    providers: [SpotifyService]
})
export class SearchBarComponent {
    searchString: string;
    results: string[];
    artists: any[]=[];
    loading: boolean;
    tracks: any []=[];
    constructor(private _spotifyService:SpotifyService){
    }

          search(query){
            console.log(query);
            this._spotifyService.getArtists( query )
                  .subscribe( (data: any) => {
                    this.artists = data;
                    console.log(this.artists);
                  });
          }
}

这是模板

<div class="container">
  <input #query id="inputbar" type="text" (keyup)="search(query.value)" class="form-control" placeholder="Search...">
  <div class="search"></div>
</div>

您的错误非常简单,您正在尝试迭代一个字符串。

当你使用switchMap时,你可以在管道内使用catchError处理错误,如下所示:

getArtists(query: string) {
  return this.searchMusic(`search?q=${query}&type=artist&limit=15`).pipe(
    switchMap(data => data["artists"].items),
    catchError(_ => of("Error!"))
  );
}

但是,您需要在订阅时再次处理它。所以我建议你让你的getArtists完好无损,并像这样处理订阅方法中的错误:

  this.spotifyService.getArtists('eeqw')
    .subscribe((data: any) => {
      console.log(data);
    }, (error: any) => {
      alert('Not found' + error);
  });

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