how to parse JSONArray in android

Question

I want to read this JSON lines but because it start with JSONArray i'm a little confused

 "abridged_cast": [
            {
                "name": "Jeff Bridges",
                "id": "162655890",
                "characters": [
                    "Jack Prescott"
                ]
            },
            {
                "name": "Charles Grodin",
                "id": "162662571",
                "characters": [
                    "Fred Wilson"
                ]
            },
            {
                "name": "Jessica Lange",
                "id": "162653068",
                "characters": [
                    "Dwan"
                ]
            },
            {
                "name": "John Randolph",
                "id": "162691889",
                "characters": [
                    "Capt. Ross"
                ]
            },
            {
                "name": "Rene Auberjonois",
                "id": "162718328",
                "characters": [
                    "Bagley"
                ]
            }
        ],

i just need to use the "name" and save all as one String. (the string value will be : Jeff Bridges,Charles Grodin,Jessica Lange,John Randolph,Rene Auberjonois).

this is my code:

try {
        //JSON is the JSON code above

        JSONObject jsonResponse = new JSONObject(JSON);
        JSONArray movies = jsonResponse.getJSONArray("characters");
        String hey = movies.toString();


    } catch (JSONException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }

I accidentally wrote "characters" in the code i ment to wrote "name". — digitalmidges, Jun 16 '13 at 19:45
Possible duplicate of [android how to convert json array to string array](http://stackoverflow.com/questions/6356231/android-how-to-convert-json-array-to-string-array) — Shreshth Kharbanda, Mar 29 '17 at 22:56

score 68 · Accepted Answer · answered Jun 16 '13 at 19:32

If you're after the 'name', why does your code snippet look like an attempt to get the 'characters'?

Anyways, this is no different from any other list- or array-like operation: you just need to iterate over the dataset and grab the information you're interested in. Retrieving all the names should look somewhat like this:

List<String> allNames = new ArrayList<String>();

JSONArray cast = jsonResponse.getJSONArray("abridged_cast");
for (int i=0; i<cast.length(); i++) {
    JSONObject actor = cast.getJSONObject(i);
    String name = actor.getString("name");
    allNames.add(name);
}

(typed straight into the browser, so not tested).

what about the json of this format [{"id":"1","user":"alll"},{"id":"1","user":"alll"}] — numerah, Sep 23 '14 at 21:03
@numerah: you have an array without a key, so just parse the json directly into a `JSONArray`. — MH., Sep 23 '14 at 23:32

score 9 · Answer 2 · answered Jun 16 '13 at 19:52

getJSONArray(attrname) will get you an array from the object of that given attribute name in your case what is happening is that for

{"abridged_cast":["name": blah...]}
^ its trying to search for a value "characters"

but you need to get into the array and then do a search for "characters"

try this

String json="{'abridged_cast':[{'name':'JeffBridges','id':'162655890','characters':['JackPrescott']},{'name':'CharlesGrodin','id':'162662571','characters':['FredWilson']},{'name':'JessicaLange','id':'162653068','characters':['Dwan']},{'name':'JohnRandolph','id':'162691889','characters':['Capt.Ross']},{'name':'ReneAuberjonois','id':'162718328','characters':['Bagley']}]}";

    JSONObject jsonResponse;
    try {
        ArrayList<String> temp = new ArrayList<String>();
        jsonResponse = new JSONObject(json);
        JSONArray movies = jsonResponse.getJSONArray("abridged_cast");
        for(int i=0;i<movies.length();i++){
            JSONObject movie = movies.getJSONObject(i);
            JSONArray characters = movie.getJSONArray("characters");
            for(int j=0;j<characters.length();j++){
                temp.add(characters.getString(j));
            }
        }
        Toast.makeText(this, "Json: "+temp, Toast.LENGTH_LONG).show();
    } catch (JSONException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }

checked it :)

score 1 · Answer 3 · answered Dec 03 '14 at 07:25

Here is a better way for doing it. Hope this helps

protected void onPostExecute(String result) {
                Log.v(TAG + " result);


                if (!result.equals("")) {

                    // Set up variables for API Call
                    ArrayList<String> list = new ArrayList<String>();

                    try {
                        JSONArray jsonArray = new JSONArray(result);

                        for (int i = 0; i < jsonArray.length(); i++) {

                            list.add(jsonArray.get(i).toString());

                        }//end for
                    } catch (JSONException e) {
                        Log.e(TAG, "onPostExecute > Try > JSONException => " + e);
                        e.printStackTrace();
                    }


                    adapter = new ArrayAdapter<String>(ListViewData.this, android.R.layout.simple_list_item_1, android.R.id.text1, list);
                    listView.setAdapter(adapter);
                    listView.setOnItemClickListener(new OnItemClickListener() {
                        @Override
                        public void onItemClick(AdapterView<?> parent, View view, int position, long id) {

                            // ListView Clicked item index
                            int itemPosition = position;

                            // ListView Clicked item value
                            String itemValue = (String) listView.getItemAtPosition(position);

                            // Show Alert
                            Toast.makeText( ListViewData.this, "Position :" + itemPosition + "  ListItem : " + itemValue, Toast.LENGTH_LONG).show();
                        }
                    });

                    adapter.notifyDataSetChanged();
...

how to parse JSONArray in android

3 Answers3

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