Android Json parsing with complicated arrays

Question

[
  {
    "countries": [
      {
        "id": 1,
        "country": "India"
      },
      {
        "id": 2,
        "country": "Australia"
      },
      {
        "id": 3,
        "country": "Srilanka"
      },
      {
        "id": 4,
        "country": "Pakistan"
      },
      {
        "id": 5,
        "country": "Switzerland"
      }
    ]
  }
]

How to parse this Json Response in Android. i am not getting any proper solution. please help me.

You need to have any name for the array.You have not mentioned it. — Harish
– Harish, Commented Aug 13, 2012 at 4:29

Kumar Vivek Mitra · Accepted Answer · 2012-08-13 04:16:21Z

4

Do it like this...

JSONArray arr = locs.getJSONArray("countries");


for (int i = 0; i < arr.length(); ++i) {
    JSONObject rec = arr.getJSONObject(i);
    int id = rec.getInt("id");
    String con = rec.getString("country");
    // ...
}

answered Aug 13, 2012 at 4:16

Kumar Vivek Mitra

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Comments

koti · Accepted Answer · 2012-08-13 04:17:26Z

1

   JSONArray jArr=new JSONArray(response);
    for(int i=0;i<jArr.length;i++)
    {
    id[]=jArr.getJSONObject(i).getInt("id");
    con[]=jArr.getJSONObject(i).getString("country");
    }

answered Aug 13, 2012 at 4:17

koti

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Comments

Nicholas Albion · Accepted Answer · 2012-08-13 04:15:01Z

0

http://developer.android.com/reference/org/json/JSONTokener.html

I've also seen Jackson used in Android apps.

answered Aug 13, 2012 at 4:15

Nicholas Albion

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Comments

Harish · Accepted Answer · 2012-08-13 04:57:39Z

0

 JSONArray countriesobj = new JSONArray();   
 JSONObject json;
            try {
                JSONObject jObject = new JSONObject(response);
                json = jObject;

                countriesobj = json.getJSONArray("countries");

                country = new String[countriesobj.length()];
                for (int i = 0; i < countriesobj.length(); i++) {

                    JSONObject e = countriesobj.getJSONObject(i);

                    country[i] = e.getString("country");

                }

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

            }

answered Aug 13, 2012 at 4:57

Harish

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