get JSON value for URL and send to another intent

Question

i am trying to fetch the json value for api. What my api is showing json as below

{
    "resultCode": 0,
    "resultMessage": "Login Success",
    "data": {
        "userid": "2",
        "username": "abc",
        "email": "mrabc@abc.com",
        "access_token": "Bw3jvwGHb4Iq4"
    }
}

how can i get the value of data??

Have you checked all answers? – Piyush Aug 06 '14 at 06:15 — Piyush, Aug 06 '14 at 06:15
thank you guyz,.. u save my day – CHIM CHIM Aug 06 '14 at 06:25 — CHIM CHIM, Aug 06 '14 at 06:25

score 2 · Answer 1 · answered Aug 06 '14 at 06:08

First convert your json response from server to a string then try like this:

JSONObject rootObject= new JSONObject(strFromServer);
JSONObject dataObject= rootObject.getJSONObject("data");
String userID = dataObject.getString("userid");
String username= dataObject.getString("username");
and so on.....

Aniruddha · Answer 2 · 2014-08-06T06:09:57.900

0

 JSONObject jobj = new JSONObject (str);

 String resultCode = jobj .getString("resultCode");
 String resultMessage = jobj .getString("resultMessage");
 JSONObject jo = jobj.getJSONObject(data);

 String userId= jo .getString("userid");
 String userName= jo .getString("username");
 String email= jo .getString("email");

edited Aug 06 '14 at 06:09

answered Aug 06 '14 at 06:03

Aniruddha

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there is no array in OP JSON "Data" is JSON object – MilapTank Aug 06 '14 at 06:05

score 0 · Answer 3 · answered Aug 06 '14 at 06:07

Every {} represents a JSON Object.

Hence you first get the JSONObject from the String : JSONObject obj = new JSONObject(jsonString);

Now, data contains another JSON Object since the contents are enclosed in {}. So, JSONObject dataObject = obj.getJSONObject("data");

Now, to get String values from the dataObject,

dataObject.getString("userId");
dataObject.getString("username"); 
...

score 0 · Answer 4 · answered Aug 06 '14 at 06:08

Use this.

 JSONObject jobj = new JSONObject (str);

 String resultCode = jobj.getString("resultCode");
 String resultMessage = jobj .getString("resultMessage");
 JSONObject jobj1 = jobj.getJSONObject("data");

 String userid= jobj1.getString("userid");
 String username= jobj1.getString("username");
 String email= jobj1.getString("email");

Pragnesh Ghoda シ · Accepted Answer · 2014-08-06T06:16:52.003

Try This...

String result = "your json string";

try {

    /* converting String to Json Object.. */
    JSONObject json_result = new JSONObject(result); 

    /* getting object values */
    String  resultCode = json_result.getJSONObject("resultCode"); 
    String  resultMessage= json_result.getJSONObject("resultMessage");

    /* creating new Object  for data */
    JSONObject json_data = jsondata.getJSONObject("data"); 

    /*fetching values from data */
    String  userid= json_data .getJSONObject("userid"); 
    String  username= json_data .getJSONObject("username");
    String  email= json_data .getJSONObject("email");
    String  access_token= json_data .getJSONObject("access_token");
} catch (JSONException e) {
    e.printStackTrace();
}

Note : Always Use Try-Catch block For converting JSON Values to string..it helps you to handle Exception and stops your application from crashing...

get JSON value for URL and send to another intent

5 Answers5