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Friday, November 21, 2014

Jquery AJAX post example with Php CodeIgniter framework

Jquery AJAX post example with Php CodeIgniter framework
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<HTML>
<HEAD>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.7.2/jquery.min.js"></script>
<script>
function makeAjaxCall(){
 $.ajax({
  type: "post",
  cache: false,   
  data: $('#userForm').serialize(),
  success: function(json){     
  try
   var obj = jQuery.parseJSON(json);
   alert( obj['STATUS']);
      
    
  }catch(e) { 
   alert('Exception while request..');
  
  },
  error: function(){     
   alert('Error while request..');
  }
 });
}
</script>
</HEAD>
<BODY>
<form name="userForm" id="userForm" action="">
<table border="1">
 <tr>
  <td valign="top" align="left"
  Username:- <input type="text" name="userName" id="userName" value="">
  </td>
 </tr>
 <tr>
  <td valign="top" align="left"
  Password :- <input type="password" name="userPassword" id="userPassword" value="">
  </td>
 </tr>
 <tr>
  <td>
   <input type="button" onclick="javascript:makeAjaxCall();" value="Submit"/>
  </td>
 </tr>
</table>
</form>
 </BODY>
</HTML>
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//UserController.php
<?php if ( ! defined('BASEPATH')) exit('No direct script access allowed');
 
class UserController extends CI_Controller {
  
 public function verifyUser() {
  $userName =  $_POST['userName'];
  $userPassword =  $_POST['userPassword'];
  $status = array("STATUS"=>"false");
  if($userName=='admin' && $userPassword=='admin'){
   $status = array("STATUS"=>"true");
  }
  echo json_encode ($status) ;
 }
}

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