Angular 如何仅从json响应中获取一些信息？

Openweatherapi返回的json如下：

{"coord":{"lon":-0.13,"lat":51.51},"weather":[{"id":300,"main":"Drizzle","description":"light intensity drizzle","icon":"09d"}],"base":"stations","main":{"temp":280.32,"pressure":1012,"humidity":81,"temp_min":279.15,"temp_max":281.15},"visibility":10000,"wind":{"speed":4.1,"deg":80},"clouds":{"all":90},"dt":1485789600,"sys":{"type":1,"id":5091,"message":0.0103,"country":"GB","sunrise":1485762037,"sunset":1485794875},"id":2643743,"name":"London","cod":200}

export class WeatherInfo{
    constructor(
                public name: string, 
                public description: string,
                public temperature: number){
                      this.name = name;
                      this.description = description;
                      this.temperature = temperature;


  } 

但我只需要这类课程的信息：

{"coord":{"lon":-0.13,"lat":51.51},"weather":[{"id":300,"main":"Drizzle","description":"light intensity drizzle","icon":"09d"}],"base":"stations","main":{"temp":280.32,"pressure":1012,"humidity":81,"temp_min":279.15,"temp_max":281.15},"visibility":10000,"wind":{"speed":4.1,"deg":80},"clouds":{"all":90},"dt":1485789600,"sys":{"type":1,"id":5091,"message":0.0103,"country":"GB","sunrise":1485762037,"sunset":1485794875},"id":2643743,"name":"London","cod":200}

export class WeatherInfo{
    constructor(
                public name: string, 
                public description: string,
                public temperature: number){
                      this.name = name;
                      this.description = description;
                      this.temperature = temperature;


  } 

}

这是我的api服务：

import { Injectable } from '@angular/core';
import { HttpClient, HttpParams } from '@angular/common/http';
import {environment} from '../environments/environment';
import { Observable } from 'rxjs';

@Injectable({
  providedIn: 'root'
})
export class ApiService {

  constructor(private http: HttpClient) { }

  getWeatherByCity(city): Observable<any>{
    const httpParams: HttpParams = new HttpParams().set('q',city).set('appid',environment.ApiKey).set('units',environment.ApiUnits);

    return this.http.get(environment.ApiUrl, {params: httpParams});

  }

  private handleError (error: any) {
    let errMsg: string;
      errMsg = error.message ? error.message : error.toString();

    console.error(errMsg);
    return Observable.throw(errMsg);
  }
}

---

我的目标是获取json数据并将其解析为weather info对象。我只需要这个类中的信息。

您可以使用rxjs

map

方法投影api调用的返回值。在您的服务中，您通常应该：

 return this.http
      .get(environment.ApiUrl, {params: httpParams})
      .pipe(map((data: any) => 
       { 
          /** Build your custom object to return **/
          /** const customObject = { id: data.id };
          /** return customObject;
       });

有关更多信息，请参阅