Hi,
The code below always returns Fail although I send correct username and password. Can anyone spot what I do wrong?
Thanks in advance
<?php
$username = htmlspecialchars(trim($_POST['username']));
$password = htmlspecialchars(trim($_POST['password']));
if($username == 1 && $password == 1)
{
echo json_encode(array('result' => 'success', 'username' => 'myname');
}
else
{
echo json_encode(array('result' => 'fail'));
}
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<script type="text/javascript" src="jquery-1.7.1.min.js"></script>
<script type"text/javascript">
$(document).ready(function()
{
$("form#login").submit(function()
{
var text_username = $('#text_username').attr('value');
var text_password = $('#text_password').attr('value');
$.ajax({
type : "POST",
url : "login.php",
data : "username=" + text_username + "&password=" + text_password,
success : function(returned)
{
if(returned.result == 'success')
{
var username = returned.username;
$('form#login').hide();
$('div#login_success').fadeIn();
}
else
{
$('form#login').hide();
$('div#login_fail').fadeIn();
}
}
});
return false;
});
});
</script>
</head>
<body>
<form id="login" method="post">
Username : <input type="text" id="text_username" />
Password : <input type="text" id="text_password" />
<br /><br />
<button type="submit">Login</button>
</form>
<div id="login_success" style="display:none; background:green;"><p>SUCCESS</p></div>
<div id="login_fail" style="display:none; background:red;"><p>FAIL</p></div>
</body>
</html>