내 JSON을보고 내 데이터가 UI에 표시되지 않습니다.

Question

이것은 제 J-SON 구문 분석으로 어디에서 실수하는지 이해하지 못합니다. 사람은 누구나 코드를보고 내가 틀렸다는 또한 J-SON의 URL

private Weather extractFeatureFromJSON (String json_incoming){ 
    Weather sendInformation = null ; 
    try { 
     JSONArray jsonArray = new JSONArray(json_incoming); 
     JSONObject jsonObject = jsonArray.getJSONObject(0); 

     long timeReceived = jsonObject.getLong("EpochTime"); 
     String weatherStatusReceived = jsonObject.getString("WeatherText"); 
     boolean DayOrNightReceived = jsonObject.getBoolean("IsDayTime"); 
     JSONObject fetchTemperature = new JSONObject("Temperature"); 
     JSONObject dive_deep_t = fetchTemperature.getJSONObject("Metric"); 
     double tempReceived = dive_deep_t.getDouble("Value"); 
     JSONObject fetch_RF_Temperature = new JSONObject("RealFeelTemperature"); 
     JSONObject dive_deep_t1 = fetch_RF_Temperature.getJSONObject("Metric"); 
     double RF_tempReceived = dive_deep_t1.getDouble("Value"); 
     JSONObject fetch_wind = new JSONObject("Wind"); 
     JSONObject fetch_wind_direction = fetch_wind.getJSONObject("Direction"); 
     int directionDegreeReceived = fetch_wind_direction.getInt("Degrees"); 
     String in_which_direction = fetch_wind_direction.getString("Localized"); 
     JSONObject fetch_wind_speed = fetch_wind.getJSONObject("Speed"); 
     JSONObject fetch_wind_speed_in_metrics = fetch_wind_speed.getJSONObject("Metric"); 
     double speedReceived = fetch_wind_speed_in_metrics.getDouble("Value"); 

     int i = Integer.parseInt(in_which_direction); 
     int directionAndInWhich = directionDegreeReceived + i; 

     return new Weather(tempReceived , RF_tempReceived , DayOrNightReceived , weatherStatusReceived ,timeReceived , speedReceived , directionAndInWhich); 
    } catch (JSONException e) { 
     e.printStackTrace(); 
    } 
    return null ; 
} 

http://dataservice.accuweather.com/currentconditions/v1/257072?apikey=JTgPZ8wN9VUy07GaOODeZfZ3sAM12irH&language=en-us&details=true

Answer 1

대신 new JSONObject

을 만드는 jsonObject에서 getJSONObject 필요를 제공하고있는 중이 야 무엇을 말해주십시오

얻을 수 Temperature, RealFeelTemperature, Wind 사용해야하는 객체

jsonObject.getJSONObject("Temperature"); 
jsonObject.getJSONObject("RealFeelTemperature"); 
jsonObject.getJSONObject("Wind"); 

는 JSONObject

new JSONObject("Temperature"); 
new JSONObject("RealFeelTemperature"); 
new JSONObject("Wind");