Android Json对象放置了一个没有键的数组Json

Android Json对象放置了一个没有键的数组Json,android,json,android-studio,android-sdk-tools,android-json,Android,Json,Android Studio,Android Sdk Tools,Android Json,我需要用Esstructure创建一个json { "label": "any description", "location":[25.7752965,-100.2636682] } 带数组的Json对象,不带键、值(位置) 我试着用 String[] _location = new String[2]; _location[0] = String.valueOf(latLng.latitud

我需要用Esstructure创建一个json

    {
      "label": "any description",
      "location":[25.7752965,-100.2636682]
    }
带数组的Json对象,不带键、值(位置)

我试着用

String[] _location = new String[2];
_location[0] = String.valueOf(latLng.latitude);
_location[1] = String.valueOf(latLng.longitude);

JSONObject params = new JSONObject();
params.put("location",_location);
params.put("label",_label);
另一个尝试使用:

       JsonObject obj = new JsonObject();
       JsonArray array_location = new JsonArray();
       array_location.add(currentLocation.latitude);
       array_location.add(currentLocation.longitude);
       JSONObject params = new JSONObject();
       params.put("location",_location);
       params.put("label",_label);
但是,当我需要一个值为数组double的键“location”时,结果会给我一个带字符串值的json

 {
     "label": "any description",
     "location":"[25.7752965,-100.2636682]"
  }

解决方案

JSONObject params = new JSONObject();
params.put("label", "any description");

JSONArray locationJsonArray = new JSONArray();
locationJsonArray.put(25.775296); // Replace by your latitude
locationJsonArray.put(-100.2636682); // Replace by your longitude
params.put("location", locationJsonArray);

// For debug purpose
Log.i("DEBUG", params.toString(4));
结果

{
  "label": "any description",
  "location": [
    25.775296,
    -100.2636682
  ]
}