JSON String in android how to get the values

Question

I have this JSON string in my android app

[{"tel1":"999","tel2":"0790000000","tel3":"","interval":"60","deleteLocal":"1","id":"2"}]

How do I parse this into a JsonArray and then get the values, for example, tel1?

If you're trying to do it with PHP, use json_decode() (with the second parameter set to TRUE): php.net/json_decode — Amal
– Amal, Commented Jan 23, 2014 at 12:26
@AmalMurali, OP wants to traverse the JSON generated from PHP into his Android App. — Shankar Narayana Damodaran
– Shankar Narayana Damodaran, Commented Jan 23, 2014 at 12:28
@ShankarDamodaran: Ah, I see. I wasn't sure about that - which is why added "If you're trying to do it with PHP" at the beginning. — Amal
– Amal, Commented Jan 23, 2014 at 12:30

keen · Accepted Answer · 2014-01-23 12:57:01Z

2

try
    {
       JSONArray jArray= new JSONArray(output);
       JSONObject jsonObject=jsonArray.getJSONObject(0);
       String tel= jsonObject.getString("tel1");
    }catch(Exception e)
    {
     //error parsing response.
    }

edited Jan 23, 2014 at 12:57

answered Jan 23, 2014 at 12:35

keen

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2 Comments

Chirag Ghori Over a year ago

if jArray having no data means length = 0 then what happen at line JSONObject jsonObject=jsonArray.getJSONObject(0);

keen Over a year ago

you can check just before 2nd line if(jArray.length()) then continue parsing.

Nambi · Accepted Answer · 2014-01-23 12:34:37Z

1

Do like This

{- represents JsonObject

[- represents JsonArray

  try
        {
        JSONArray jArray= new JSONArray(output);
               JSONObject menuObject = JSONArray.getJSONObject(0);  
               String tel= menuObject.getString("tel1");
        }catch(Exception e)
        {

Log.e("d",e.getMessage());
        }

edited Jan 23, 2014 at 12:34

answered Jan 23, 2014 at 12:29

Nambi

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Comments

Chirag Ghori · Accepted Answer · 2014-01-23 12:37:54Z

1

Try with this

try {
        JSONArray jsonArray = new JSONArray(responsestring);
        for (int i = 0; i < jsonArray.length(); i++) {

            JSONObject jobj = jsonArray.getJSONObject(i);
            String strtel1 = jobj.getString("tel1");

            Log.i("value tel1 : ", strtel1 + "");
        }
    } catch (JSONException e) {
        // TODO: handle exception
    }

answered Jan 23, 2014 at 12:37

Chirag Ghori

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3 Comments

Tom S Over a year ago

I get this error Cannot make a static reference to the non-static method getJSONObject(int) from the type JSONArray

Chirag Ghori Over a year ago

sorry i made a mistake now check my answer ...use this JSONObject jobj = jsonArray.getJSONObject(i);

Chirag Ghori Over a year ago

@TomS : Heyy in your accepted answer if jArray having no data means length = 0 then what happen at line JSONObject jsonObject=jsonArray.getJSONObject(0); ???

umair · Accepted Answer · 2014-01-23 12:39:37Z

1

Try This.

JSONArray arr = new JSONArray(result);
JSONObject jObj = arr.getJSONObject(0);
String tel1 = jObj.getString("tell");

answered Jan 23, 2014 at 12:39

umair

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Comments

Houcine · Accepted Answer · 2014-01-23 14:10:41Z

0

try this :

try {
    JSONArray jsonArray= new JSONArray(yourString);
    if(jsonArray != null && jsonArray.length() > 0) {
        for(int i=0; i< jsonArray.length(); i++) {
            JSONObject json = jsonArray.getJSONObject(i);
            if(json != null) {
                String tel1 = json.optString("tel1", "default_value");
                Log.d("PARSE", "tel1  : "+tel1);
            }
        }
    }
}catch(Exception e)
{
}

edited Jan 23, 2014 at 14:10

answered Jan 23, 2014 at 12:36

Houcine

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