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POpassing array of data using JSON in android
 

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  1. POpassing array of data using JSON in android
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    <p>I need to pass data from MYSQL database to android application by creating JSON object. I am able to pass only the first value from MYSQL database. How to pass all the values together to android application. (My database contains Latitude and Longitude values of more than 10 locations.)</p> <p>Following is my code for passing only the first row value from database to android app.</p> <pre><code> try { HttpClient client = new DefaultHttpClient(); URI website = new URI("http://192.168.1.15/latlonret1.php"); HttpGet request = new HttpGet(); request.setURI(website); HttpResponse response = client.execute(request); in = new BufferedReader(new InputStreamReader(response .getEntity().getContent())); StringBuffer sb = new StringBuffer(""); String l = ""; String nl = System.getProperty("line.separator"); while ((l = in.readLine()) != null) { sb.append(l + nl); } in.close(); data = sb.toString(); // return data; } catch (Exception e) { Log.e("log_tag", "Error converting result " + e.toString()); } try { String returnString; String returnString1; JSONArray jArray = new JSONArray(data); for (int i = 0; i &lt; jArray.length(); i++) { JSONObject json_data = jArray.getJSONObject(i); JSONObject json_data1 = jArray.getJSONObject(i); returnString=json_data.getString("lat") + "\n"; returnString1=json_data1.getString("lon") + "\n"; System.out.println(returnString); System.out.println(returnString1); Intent viewIntent =new Intent(Androidmap.this,Mapview.class); Bundle bundle = new Bundle(); bundle.putString("stuff", returnString); viewIntent.putExtras(bundle); Bundle bundle1 = new Bundle(); bundle1.putString("stuff1", returnString1); viewIntent.putExtras(bundle1); startActivity(viewIntent); } </code></pre> <p>This is my Server side coding :</p> <pre><code>&lt;?php ob_start(); $host="localhost"; // Host name $username=""; // Mysql username $password=""; // Mysql password $db_name="test"; // Database name $tbl_name="manu"; // Table name mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); $sql = "select lat , lon from manu "; $result=mysql_query($sql); if(! $result ) { die('Could not get data: ' . mysql_error()); } while($row=mysql_fetch_array($result , MYSQL_ASSOC)) { $output[]=$row; } echo json_encode($output); mysql_close(); ?&gt; </code></pre>
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    1. USSahil Mahajan Mj
    1. USuser1844654
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    1. COHow do you created serevr side using PHP ?
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      1. USAnoop
    2. COCan you post the JSON? It's hard to say what's going on without seeing that...
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      1. USLou Morda
    3. COI have updated my question. Please take a look
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