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  1. PO
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    <p>Please note that i have used Strict Policy. You continue using Async Task and put my code in there. Here is my Code related to your requirement . </p> <p>JsonParser.Java</p> <pre><code> package com.example.test; import android.os.StrictMode; import android.util.Log; import org.apache.http.HttpEntity; import org.apache.http.HttpResponse; import org.apache.http.client.ClientProtocolException; import org.apache.http.client.methods.HttpPost; import org.apache.http.impl.client.DefaultHttpClient; import org.json.JSONArray; import org.json.JSONException; import org.json.JSONObject; import java.io.BufferedReader; import java.io.IOException; import java.io.InputStream; import java.io.InputStreamReader; import java.io.UnsupportedEncodingException; public class JSONParser { static InputStream is = null; static JSONObject jObj = null; static String json = ""; static JSONArray jArray = null; // constructor public JSONParser() { } public JSONArray getJSONFromUrl(String url) { // Making HTTP request try { StrictMode.ThreadPolicy policy = new StrictMode.ThreadPolicy.Builder().permitAll().build(); StrictMode.setThreadPolicy(policy); // defaultHttpClient DefaultHttpClient httpClient = new DefaultHttpClient(); HttpPost httpPost = new HttpPost(url); HttpResponse httpResponse = httpClient.execute(httpPost); HttpEntity httpEntity = httpResponse.getEntity(); is = httpEntity.getContent(); } catch (UnsupportedEncodingException e) { e.printStackTrace(); } catch (ClientProtocolException e) { e.printStackTrace(); } catch (IOException e) { e.printStackTrace(); } try { BufferedReader reader = new BufferedReader(new InputStreamReader( is, "iso-8859-1"), 8); StringBuilder sb = new StringBuilder(); String line = null; while ((line = reader.readLine()) != null) { sb.append(line + "\n"); } is.close(); json = sb.toString(); } catch (Exception e) { Log.e("Buffer Error", "Error converting result " + e.toString()); } // try parse the string to a JSON object try { jArray = new JSONArray(json); // jObj = new JSONObject(json); } catch (JSONException e) { Log.e("JSON Parser", "Error parsing data " + e.toString()); } // return JSON String return jArray; } } </code></pre> <p>MainActivity.java</p> <pre><code> package com.example.test; import java.io.BufferedReader; import java.io.IOException; import java.io.InputStream; import java.io.InputStreamReader; import java.io.UnsupportedEncodingException; import org.apache.http.HttpEntity; import org.apache.http.HttpResponse; import org.apache.http.client.ClientProtocolException; import org.apache.http.client.methods.HttpPost; import org.apache.http.impl.client.DefaultHttpClient; import org.json.JSONArray; import org.json.JSONException; import org.json.JSONObject; import android.app.Activity; import android.os.Bundle; import android.os.StrictMode; import android.util.Log; import android.view.Menu; public class MainActivity extends Activity { static InputStream is = null; static JSONObject jObj = null; static String json = ""; @Override protected void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.activity_main); // Making HTTP request String url = "http://api.worldbank.org/countries/ir?format=json"; JSONParser jParser = new JSONParser(); // Getting JSON from URL JSONArray json = jParser.getJSONFromUrl(url); try { JSONArray json2 = json.getJSONArray(1); // Your second Array in your json for(int i=0;i&lt;json2.length();i++){ JSONObject c = json2.getJSONObject(i); // Storing JSON item in a Variable String name = c.optString("name",""); String capitalCity = c.optString("capitalCity",""); } } catch (JSONException e) { e.printStackTrace(); } } @Override public boolean onCreateOptionsMenu(Menu menu) { // Inflate the menu; this adds items to the action bar if it is present. getMenuInflater().inflate(R.menu.main, menu); return true; } } </code></pre>
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    1. POHow to get the data in a array from URL
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      1. PTQuestion
    1. PTAnswer
    1. This table or related slice is empty.
    1. USRahul Gupta
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    1. COThanks, but I could not try it as there is an error in `JSONArray json2 = json.getJSONArray(1);` there should be a string instead of 1 in this line. Any idea?
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      1. USuser2977338
    2. COYou have an array at 1st position in your json. This line gets this. This code works fine. I have run it like many times now. There shouldn't be any String there as it takes in int value. Check our json. You have a JSON Array in your JSON at 1st position. This line access that part of your json. Debug your code.,.,
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      1. USRahul Gupta
 

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