Question

我的for (int i = 0; i <= response.length(); i++) { JSONObject lightObject = response.getJSONObject(String.valueOf(i)); System.out.println("OBJECT " + lightObject); // Use try/catch because of empty or unknown key try { description = lightObject.getString("type"); } catch (Exception e) {} try { bri = lightObjectState.getInt("bri"); } catch (Exception e) {} try { sat = lightObjectState.getInt("sat"); } catch (Exception e) {} try { hue = lightObjectState.getInt("hue"); } catch (Exception e) {} Light light = new Light(String.valueOf(i),lightObject.getString("name"), description, on, bri, sat, hue); mLightList.add(light); mLightAdapter.notifyDataSetChanged(); }包含其他对象：

no value of 0

飞利浦Hue Brigde JSON的响应包含所有连接的灯光。我使用Volley进行http连接。我创建了一个Light类，我存储了像state，hue，bri等数据。我通过在Volley的onResponse中使用for循环来填充Light类：

key

for循环的问题在于，当Light id以大于零开始时，它将使我的Android应用程序崩溃，因为有JSONObject。我的问题是如何在不知道包含具有一个灯光的所有属性的对象的{ "4" : { "state": [], "hue": 224 } "5" : { "state": [], "hue": 224 } "6" : { "state": [], "hue": 224 } }名称的情况下获取响应的第一个对象，因此当{ "data": [ { "target": "test-series-0", "datapoints": [ [ 22.504392773143504, 1.476693264195e+12 ], [ 22.719552781746028, 1.476693301825e+12 ] ] } ] }如下所示时，应用程序将不会崩溃，但只显示4到6的光：

{
  "data": [
    {
      "target": "test-series-0",
      "datapoints": [
        [
          1.476693264195e+12
          22.504392773143504,
        ],
        [
          1.476693301825e+12
          22.719552781746028,
        ]
      ]
    }
  ]
}

如果还有任何问题，请告诉我。感谢。

Answer 1

    // use keys() iterator, you don't need to know what keys are there/missing
    Iterator<String> iter = response.keys();
    while (iter.hasNext()) {
        String key = iter.next();
        JSONObject lightObject = response.getJSONObject(key); 
        System.out.println("key: " + key + ", OBJECT " + lightObject);

        // you can check if the object has a key before you access it
        if (lightObject.has("bri")) {
            bri = lightObject.getInt("bri");
        }

        if (lightObject.has("sat")) {
            sat = lightObject.getInt("sat");
        }

        if (lightObject.has("hue")) {
            hue = lightObject.getInt("hue");
        }

        if (lightObject.has("name")) {
            name = lightObject.getString("name")
        }

        Light light = new Light(key, name, description, on, bri, sat, hue);
        mLightList.add(light);
    }

    mLightAdapter.notifyDataSetChanged();

Answer 2

尝试使用对象键：

 Iterator<String> keys = jsonObject.keys();
 if( keys.hasNext() ){
     String key = (String)keys.next(); // First key in your json object
  }

Answer 3

假设您有一个名为Bulb的课程

public class Bulb {

    int[] state;
    int hue;

}

然后你可以使用像GSON

这样的JSON序列化器

然后你可以将你的回复映射到地图（如HashMap），就像这样

Type typeOfHashMap = new TypeToken<Map<String, Bulb>>() { }.getType();

Map<String,Bulb> bulbMap = gson.fromJson(json, typeOfHashMap);

然后你可以用这样的键遍历地图

Iterator bulbIterator = bulbMap.entrySet().iterator();

  while (bulbIterator.hasNext()) {

    Map.Entry pair = (Map.Entry)bulbIterator.next();

    String pKey = (String) pair.getKey();

    Bulb bulb = (Bulb) pair.getValue();

    // Do something here with the bulb
}

如何在不知道名称的情况下获取JSONObject中的第一个对象？

3 个答案: