Question

    public void HelloWord {
       for (int i = 0; i < 20; i++) {
             Log.d("Great");
       }
    }

The code above doesn’t work why?
I try to get value name
Does anybody know where is the problem?

org.json.JSONException: Value …Content of link… at
org.json.JSON.typeMismatch at org.json.JSONArray.

Answer 1

I think that you are missing the way that JSON works. Anything in {} is an Object, while [] designates an array. So the root of the twitter feed is a JSONObject, NOT a JSONArray:

Try something more like this:

JSONObject obj = new JSONObject(mStringBuilder.toString());
JSONObject trends = obj.getJSONObject("trends");
JSONArray today = trends.getJSONArray("2012-04-10");
for (int i = 0; i < today.length(); i++) {
  JSONObject tag = today.getJSONObject(i);
  String name = tag.getString("name");
  // do whatever with name
}

Much easier, and its clearer how it works. JSONObjects are dictionaries, with a simple mapping between keys and values – each Object ({}) can contain either more objects, or arrays ([]) which can contain either simple integers or more objects