需要在用户添加任何过滤器之前将所有项目加载到 Observable<Hero[]>

Question

如何更改 Angular 2 Tour of Heroes 搜索组件 (https://angular.io/generated/live-examples/toh-pt6/eplnkr.html)，以便它在初始化时带来所有项目（在页面加载时显示所有英雄），并且当提供过滤器时，它会向服务发出新请求过滤后的结果放入 heros 变量中？

import { Component, OnInit } from '@angular/core';
import { Router }            from '@angular/router';

import { Observable }        from 'rxjs/Observable';
import { Subject }           from 'rxjs/Subject';

// Observable class extensions
import 'rxjs/add/observable/of';

// Observable operators
import 'rxjs/add/operator/catch';
import 'rxjs/add/operator/debounceTime';
import 'rxjs/add/operator/distinctUntilChanged';

import { HeroSearchService } from './hero-search.service';
import { Hero } from './hero';

@Component({
  selector: 'hero-search',
  templateUrl: './hero-search.component.html',
  styleUrls: [ './hero-search.component.css' ],
  providers: [HeroSearchService]
})
export class HeroSearchComponent implements OnInit {
  heroes: Observable<Hero[]>;
  private searchTerms = new Subject<string>();

  constructor(
    private heroSearchService: HeroSearchService,
    private router: Router) {}

  // Push a search term into the observable stream.
  search(term: string): void {
    this.searchTerms.next(term);
  }

  ngOnInit(): void {
    this.heroes = this.searchTerms
      .debounceTime(300)        // wait 300ms after each keystroke before considering the term
      .distinctUntilChanged()   // ignore if next search term is same as previous
      .switchMap(term => term   // switch to new observable each time the term changes
        // return the http search observable
        ? this.heroSearchService.search(term)
        // or the observable of empty heroes if there was no search term
        : Observable.of<Hero[]>([]))
      .catch(error => {
        // TODO: add real error handling
        console.log(error);
        return Observable.of<Hero[]>([]);
      });
  }

  gotoDetail(hero: Hero): void {
    let link = ['/detail', hero.id];
    this.router.navigate(link);
  }
}

目前只是在提供搜索词后发送请求。

Answer 1

你好_

基本上这是您可以进行更改以使其正常工作的主要地方：

.switchMap(
    term => term    
    ? this.heroSearchService.search(term)
    : Observable.of<Hero[]>([]))  // <----- HERE if term is empty string you want to return all Heroes instead of empty collection 

要实现这一点，您可以从 hero.service.ts 注入 HeroService，该方法具有返回所有英雄的方法 getHeroes()。

所以现在我们可以把上面的代码改成这样：

// First don't forget to inject HeroService and don't forget to import it too
constructor(
    private heroSearchService: HeroSearchService,
    private heroService: HeroService,
    private router: Router) {}

// Then the ngOnInit() will look like this
ngOnInit(): void {
    this.heroes = this.searchTerms
      .debounceTime(300)        
      .distinctUntilChanged()  
      .switchMap(term => term  
        ? this.heroSearchService.search(term)
        : this.heroService.getHeroes()) // <--- HERE if term is empty return all heroes
      .catch(error => {
        console.log(error);
        return Observable.of<Hero[]>([]);
      });

      // Wait for 100 ms before loading all heroes ...
      setTimeout(() => {
        this.search('');
      }, 100); 
} 

我尝试添加对 this.search(""); 的调用在 ngOnInit() 内部

是的，我们需要这个来加载所有英雄，但是稍微延迟一下，否则什么都不会显示。

如果一切都清楚，请告诉我:)