json response get values in android

Question

I am new to android and struggling to get the values of the json object. Can someone help me out?

the json returned from the server is {"status":"active"}

I'm using Android Asynchronous Http Client library..

public void onSuccess(int statusCode, Header[] headers, JSONObject response) {

            try {
                JSONArray j = new JSONArray(response);
                t.setText(j.getJSONObject(0).getString('status'));//this doesn't set the text to the status
            } catch (JSONException e) {
                Log.e("MYAPP", "unexpected JSON exception", e);
            }
        }

It seems you need to understand the basics of JSON data. Like {} represents JSON object, [] represents Json array. — Paresh Mayani, Feb 22 '17 at 04:53
Please see [Why is “Can someone help me?” not an actual question?](http://meta.stackoverflow.com/q/284236) — EJoshuaS - Stand with Ukraine, Feb 22 '17 at 05:01
Possible duplicate of [How to parse JSON in Android](http://stackoverflow.com/questions/9605913/how-to-parse-json-in-android) — Nilabja, Feb 22 '17 at 07:02

Kiran Benny Joseph · Answer 1 · 2017-02-22T05:06:53.760

you can use

JSONObject jsonObject = new JSONObject(response);

because {"status":"active"} is a JSONObject.

and use

 t.setText(jsonObject.getString("status"));

Full code is

public void onSuccess(int statusCode, Header[] headers, JSONObject response) {

            try {
                JSONObject jsonObject = new JSONObject(response);
                t.setText(jsonObject.getString("status"));
            } catch (JSONException e) {
                Log.e("MYAPP", "unexpected JSON exception", e);
            }
        }

As per other answers It is waste to create a new JSONObject. So use it directly

   public void onSuccess(int statusCode, Header[] headers, JSONObject response) {

                try {

                    t.setText(response.getString("status"));
                } catch (JSONException e) {
                    Log.e("MYAPP", "unexpected JSON exception", e);
                }
            }

Note: creating more instances is for bad performance

Thanks. I know yeah since I would have done obj[0].status in javascript but it didn't work here. — maxcc, Feb 22 '17 at 05:10

score 1 · Answer 2 · answered Feb 22 '17 at 05:02

1

It's already a JSON Object as incoming parameter You don't need to create external object of JSON Array or JSON Object Try This.

t.setText(response.getString("status"));

answered Feb 22 '17 at 05:02

Maaz Patel

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Zaki Pathan · Accepted Answer · 2017-02-22T05:09:12.463

0

replace this:

JSONArray j = new JSONArray(response);
t.setText(j.getJSONObject(0).getString("status"));

with:

JSONObject j = new JSONObject(response);
t.setText(j.getString("status"));

if you getting String in response you have to try this hope it helps you

EDIT:

But in your case JSONObject is coming in response So you can try directly like this:

t.setText(response.getString("status"));

edited Feb 22 '17 at 05:09

answered Feb 22 '17 at 04:52

Zaki Pathan

1,762
1
13
26

it says can't resolve toString() which library do I import? – maxcc Feb 22 '17 at 04:57
check my edited answer. try t.setText(response.getString("status")); directly in place of JSONArray j = new JSONArray(response); t.setText(j.getJSONObject(0).getString("status")); – Zaki Pathan Feb 22 '17 at 04:58
if this works mark this as a correct answer so others getting help from this. Happy to help you :) :) :) – Zaki Pathan Feb 22 '17 at 05:01

score 0 · Answer 4 · answered Feb 22 '17 at 05:00

Use direct JSONObject for getting value

 public void onSuccess(int statusCode, Header[] headers, JSONObject response) {

       try {
            t.setText(response.getString("status"));
        } catch (JSONException e) {
            Log.e("MYAPP", "unexpected JSON exception", e);
        }
    }

score 0 · Answer 5 · answered Feb 22 '17 at 05:12

It seems you are not familiar with JSON object and JSON array. I will recommend you to print your response on Log.i("response", response.toString()) then after checking response from JSON validator You perform actual stuff.

see JSON Array and JSON object also

score 0 · Answer 6 · answered Feb 22 '17 at 05:35

Use optString instead of getString,because if you are using getString and the key "status" not exit then app may be crash.

public void onSuccess(int statusCode, Header[] headers, JSONObject response) {

            try {

              t.setText(j.getJSONObject(0).optString('status'));

            } catch (JSONException e) {
                Log.e("MYAPP", "unexpected JSON exception", e);
            }
        }

json response get values in android

6 Answers6