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    <p>Tonity's solution is good. You can also use this solution.</p> <p>In my solution, there will be no any <strong>Exception</strong> fired <strong>until JSON is wrong</strong>. What I am doing is following.</p> <blockquote> <ol> <li>Search for number of ":" in string. </li> <li>If it returns 1, then we sure that there is "name" value.</li> <li>Otherwise, we need to check, whether there is "FirstName","MiddleName" or "LastName" exist in string or not.</li> </ol> </blockquote> <p>Just go through this snippet and you will find solution for your problem.</p> <pre><code>// String str = "{\"name\":\"chintan\"}"; String str = "{\"name\":{\"FirstName\":\"Chintan\",\"LastName\":\"Rathod\"}}"; try { //we will check how many ":" are there in string, if it is one, then // we are going to get "name" field. if ((str.split(":").length - 1) == 1) { Log.d("Home", "1"); JSONObject json = new JSONObject(str); Log.d("Home", "Name : " + json.get("name")); } else { Log.d("Home", "more then 2"); JSONObject jName = new JSONObject(str); JSONObject jTemp = jName.getJSONObject("name"); if (jTemp.toString().contains("FirstName")) Log.d("Home", "FirstName :" + jTemp.getString("FirstName")); if (jTemp.toString().contains("MiddleName")) Log.d("Home","MiddleName :" +jTemp.getString("MiddleName")); if (jTemp.toString().contains("LastName")) Log.d("Home", "LastName :" + jTemp.getString("LastName")); } } catch (JSONException e) { e.printStackTrace(); } </code></pre> <p><strong>Output</strong></p> <pre><code>08-06 11:52:34.060: D/Home(1439): more then 2 08-06 11:52:34.060: D/Home(1439): FirstName :Chintan 08-06 11:52:34.070: D/Home(1439): LastName :Rathod </code></pre>
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    1. POJava - Parsing JSON Response - Content is either String or JSONObject
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