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
– Paresh Mayani, Commented Feb 22, 2017 at 4:53
Please see Why is “Can someone help me?” not an actual question? — EJoshuaS - Stand with Ukraine
– EJoshuaS - Stand with Ukraine, Commented Feb 22, 2017 at 5:01

Maaz Patel · Accepted Answer · 2017-02-22 05:02:22Z

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"));

Kiran Benny Joseph · Accepted Answer · 2017-02-22 05:06:53Z

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.

Zaki Pathan · Accepted Answer · 2017-02-22 05:09:12Z

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"));

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"));
if this works mark this as a correct answer so others getting help from this. Happy to help you :) :) :)

Sasi Kumar · Accepted Answer · 2017-02-22 05:00:25Z

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); } }

Muhammad Usman · Accepted Answer · 2017-02-22 05:12:19Z

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

Zakir hussain · Accepted Answer · 2017-02-22 05:35:39Z

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); } }

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