我正在学习官方的角度教程(https://angular.io/tutorial/toh-pt4)。请问我的低水平:(
我正在尝试修改一个返回英雄列表的服务...作为rxjs的可观察对象,因为这是最有效的方法
hero.service.ts
==============
import {Injectable} from '@ angular / core';
import {Observable, of} from 'rxjs';
import {Hero} from './hero';
import {HEROES} from './mock-heroes';
import {MessageService} from './message.service';
@Injectable ({
providedIn: 'root',
})
export class HeroService {
constructor (private messageService: MessageService) {}
getHeroes (): Observable <Hero []> {
// TODO: send the message _after_ fetching the heroes
this.messageService.add ('HeroService: fetched heroes');
return of (HEROES);
}
}
mock-heroes
===========
import {Hero} from './hero';
export const HEROES: Hero [] = [
{id: 11, name: 'Mr. Nice '},
{id: 12, name: 'Narco'},
{id: 13, name: 'Bombast'},
{id: 14, name: 'Celeritas'},
{id: 15, name: 'Magneta'},
{id: 16, name: 'RubberMan'},
{id: 17, name: 'Dynama'},
{id: 18, name: 'Dr IQ'},
{id: 19, name: 'Magma'},
{id: 20, name: 'Tornado'}
];
我想提供一种在传递ID时仅返回英雄的服务
import {Injectable} from '@ angular / core';
import {Hero} from './hero';
import {HEROES} from './mock-heroes';
import {Observable, of} from 'rxjs';
import {MessageService} from './message.service';
import {map} from 'rxjs / operators';
@Injectable ({
providedIn: 'root'
})
export class HeroService {
private heroes: Observable <Hero []>;
getHeroes (): Observable <Hero []> {
this.messageService.add ('HeroService: fetched heroes');
// return of (HEROES);
this.heroes = of (HEROES);
return this.heroes;
}
getHeroeById (id: number): Observable <Hero> {
this.messageService.add ('A hero is queried with id' + id);
return this.heroes.pipe (map (x => x.id === id));
}
constructor (private messageService: MessageService) {}
/ * getHeroes (): Hero [] {
return HEROES;
} * /
//builder() { }
}
src / app / hero.service.ts中的错误(26,5):错误TS2322:类型'Observable'不能分配给类型'Observable'。 “布尔”类型不能分配给“英雄”类型。 src / app / hero.service.ts(26.40):TS2339错误:类型“英雄[]”上不存在属性“ id”。
你能帮我吗?
答案 0 :(得分:2)
您有两种选择。您可以使用Array.find()或Array.filter()
getHeroeById (id: number): Observable <Hero> {
this.messageService.add('A hero is queried with id: ' + id);
// Filter heroes to get heroById
return this.heroes.filter(x => x.id == id)[0];
}
getHeroeById (id: number): Observable <Hero> {
this.messageService.add('A hero is queried with id: ' + id);
// Find hero by Id.
return this.heroes.find(x => x.id == this.personId);
}
答案 1 :(得分:1)
您要返回英雄列表中具有匹配ID的英雄,请尝试替换
return this.heroes.pipe (map (x => x.id === id));
作者
return this.heroes.pipe (map (heroes => heroes.find( hero => hero.id === id));
答案 2 :(得分:1)
您的问题是此行:
return this.heroes.pipe (map (x => x.id === id));
您正在使用rx map运算符,该运算符对流中的项目进行操作并对其进行转换,并且在这种情况下,流项目是英雄的实际数组。您将返回该数组的某些不存在的id属性是否等于id参数(该值为布尔值)的测试结果。这就是为什么看到错误的原因。
在这种情况下,应在流中对数组使用数组查找运算符,因为要在数组中查找该项,就像这样:
return this.heroes.pipe (map (heroes => heroes.find(h => h.id === id)));
这样,您将在rx映射转换操作中返回数组查找操作的实际结果。
接下来,在调用heroes
之前,您不会初始化getHeroes()
可观察对象。这可能是故意的,但似乎您应该改为在构造函数中执行此操作,以便可以首先安全地调用getHeroeById()
。