Question

这是我的J-SON解析我不明白我在哪里弄错了。任何人请看代码并告诉我我做错了什么。我还提供了J-SON 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

您需要getJSONObject jsonObject而不是创建new JSONObject

要获取Temperature, RealFeelTemperature, Wind对象，您需要使用

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

不是新JSONObject

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

看看我的JSON，我的数据没有在UI中显示

1 个答案: