Jquery AJAX post example with Php CodeIgniter framework
<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",
url: "http://localhost/CodeIgnitorTutorial/index.php/usercontroller/verifyUser",
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>
//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) ;
}
}
