Decode JSON String in Android

Question

I have a contacts which is in the form of JSON. Now I want to decode them into String array. There are two arrays; names and phones. I'm using this code:

    String[] names;
    String[] phones;

    String test = "[{\"name\":\"A\",\"phone\":\"911\"},{\"name\":\"A1\",\"phone\":\"911\"},{\"name\":\"Abid\",\"phone\":\"371812\"}]";
    try {
        JSONArray jsonArray = new JSONArray(test);
        JSONObject jsonObject = new JSONObject(jsonArray.toString());
        Log.i("INFO", String.valueOf(jsonObject.length()));
    } catch (JSONException e) {
        e.printStackTrace();
    }

This generates an error. How can I add all names in names array and all phones in phones array. Like names[0] is assigned A which is a first name and phones[0] assigned 911 which is first phone number corresponding to first name. How can I do that, I'm new in android?

What is new JSONObject(jsonArray.toString()) supposed to achieve?! — Biffen
– Biffen, Commented Feb 21, 2017 at 5:47
I think there is objects within array So, I try to retrieve these. Like there are three objects, isn't it ? — user7423754
– user7423754, Commented Feb 21, 2017 at 5:48
you have to fetch each object by lopping through it , you can't simply get it like that — Prashant
– Prashant, Commented Feb 21, 2017 at 5:49
@PakDeveloper Yeah, so why are you trying to turn the array into one object?! — Biffen
– Biffen, Commented Feb 21, 2017 at 5:49

Prashant · Accepted Answer · 2017-02-21 06:07:58Z

3

The problem is in this line

JSONObject jsonObject = new JSONObject(jsonArray.toString());

you are trying to convert JSONArray in JSONObject , which you can't. if you want to access JSONObject inside of JSONArray you have to loop through each, or you can get specific object by it's index.

    String[] names;
    String[] phones;

    String test = "[{\"name\":\"A\",\"phone\":\"911\"},{\"name\":\"A1\",\"phone\":\"911\"},{\"name\":\"Abid\",\"phone\":\"371812\"}]";
    try {
        JSONArray jsonArray = new JSONArray(test);
        phones = names = new String[jsonArray.length()];
        for(int i = 0 ; i < jsonArray.length(); i ++){
            JSONObject jsonObject = jsonArray.getJSONObject(i);
            names[i] = jsonObject.getString("name");
            phones[i] = jsonObject.getString("phone");
            Log.i("INFO", "name : " + jsonObject.getString("name") + " , phone : " + jsonObject.getString("phone"));

        }
    } catch (JSONException e) {
        e.printStackTrace();
    }

edited Feb 21, 2017 at 6:07

answered Feb 21, 2017 at 5:48

Prashant

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1 Comment

IntelliJ Amiya Over a year ago

This answer looks perfect .

Piyush · Accepted Answer · 2017-02-21 05:52:10Z

1

You can't convert json arrya to json object like that:

try {
        JSONArray jsonArray = new JSONArray(test);
        for(int i=0; i<jsonArray.length(); i++) {
           JSONObject jsonObject = jsonArray.getJSONObject(i);
           // Your code goes here..  
        }

    } catch (JSONException e) {
        e.printStackTrace();
    }

edited Feb 21, 2017 at 5:52

answered Feb 21, 2017 at 5:48

Piyush

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Comments

Ankush Bist · Accepted Answer · 2017-02-21 05:49:15Z

here is the required code.

 ArrayList<String> names = new ArrayList<>();
    ArrayList<String> phones = new ArrayList<>();

    String test = "[{\"name\":\"A\",\"phone\":\"911\"},{\"name\":\"A1\",\"phone\":\"911\"},{\"name\":\"Abid\",\"phone\":\"371812\"}]";
    try {
        JSONArray jsonArray = new JSONArray(test);
        for (int i = 0; i < jsonArray.length(); i++) {
            JSONObject jsonObject = jsonArray.getJSONObject(i);
            String name = jsonObject.getString("name");
            String phone = jsonObject.getString("phone");
            names.add(name);
            phones.add(phone);
            Log.i("INFO", name + ", " + phone);
        }
    } catch (JSONException e) {
        e.printStackTrace();
    }

above you have used simple arrays they can store data only of fixed size. But arrayList and List can store data of variable size or you do not need to fix their size.

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