Question

我正在使用

 ResponseEntity<List<item>> res = restTemplate.exchange(
                "http://localhost:8080/page",
                HttpMethod.GET,
                null,
                new ParameterizedTypeReference<List<item>>() {});

对于我的一个应用程序，但由于我进行了休息服务，它只通过Json返回了2个项目。我的问题是，我正在尝试获取天气和api我正在使用返回：

{"coord":{"lon":-0.13,"lat":51.51},"weather":    [{"id":741,"main":"Fog","description":"fog","icon":"50n"},{"id":701,"main":"Mist","description":"mist","icon":"50n"}],"base":"stations","main":{"temp":278.47,"pressure":1012,"humidity":100,"temp_min":277.15,"temp_max":279.15},"visibility":10000,"wind":{"speed":1},"clouds":{"all":75},"dt":1479864000,"sys":{"type":1,"id":5093,"message":0.0311,"country":"GB","sunrise":1479886350,"sunset":1479916868},"id":2643743,"name":"London","cod":200}

那是很多JSON，所以我想知道我是否必须制作一个包含所有那些具有setter / getters（coord字段，id，描述，天气，名称等等）的对象的类，或者我可以只是像coord对象和天气对象一样创建一个对象，并且只有那些字段映射？

感谢您的任何建议。

编辑：

也许它与@JsonIgnoreProperties标签的ignoreUnknown = true属性有关？

Answer 1

我认为你只需要两个类来设置getter和setter类。你的天气＆＃34;变量只是List，其他是静态变量。我会这样做：

public class Weather {

  private Long id;

  private String name;

  private Long cod;

  private coordLong string;

  private coordLat string;

  priva... baseStations, cloudsAll, windSpeed, mainTemp...

  private List<WeatherList> weatherList;

  ...
  //getters setters

Answer 2

如果您想要映射所有变量，只需创建一个Model类并映射您的响应，就像这样

public class WeatherData{

private Coordinate coord;
private List<Weather> weather;

//Getters and setters setCoord(),getCoord().setWeather(),getWeather().

public class Coordinate{

    private Double lon;
    private Double lat;
    //Getters and Setters
}

public class Weather{

    private Integer id;
    private String main;
    private String description;
    private String icon;

    //Getters and setters
}

//Remaining fields

}

然后像这样映射你的json

//json is your response in String format
Gson gson = new Gson();
WeatherData weatherData = gson.fromJson(json,WeatherData.class);
//handle exception here

然后得到像这样的任何变量值

lat = weatherData.getCoord().getLat();

使用Exchange（）将Json转换为Java / Spring中的Object

2 个答案: