AJAX not passing data to PHP script?

Question

I have this code:

<form id="login" name="login" class="login-form" action="./" method="post">
      <h2>Log In</h2>
      <div class="error" style="display:none"></div>
      <input type="text" id="login-username" placeholder="Full Name" />
      <input type="password" id="login-password" placeholder="Password" />
      <button onclick="submitLogin();">Log In</button>
    </form>

    <script type="text/javascript">
$(".login").submit(function(ev){
  ev.preventDefault();
});

function submitLogin() {
    username = $(".login-username").val();
    password = $(".login-password").val();
    $.ajax({
        type: "POST",
        url: "login.php",
        datatype: "string",
        data: {'username' : username, 'password' : password},

.....

And this is my PHP script (login.php):

<?php
session_start();
require_once 'config.php';

$uName = $_POST['username'];
$pWord = md5($_POST['password']);

$query = "SELECT id, username, password, email FROM users WHERE username = '$uName' AND password = '$pWord'";
$result = mysql_query($query) or die(mysql_error());
$numrows = mysql_num_rows($result);
$row = mysql_fetch_assoc($result);


if($numrows == 1 ) {
    echo 'true';
    $_SESSION['uName'] = $row['username'];
}
else {
    echo 'false';
    }
?>

When I try to echo out $uName or $pWord its empty. Not sure whats going on. Help would be awesome. When I edit a correct username and password to $uName and $pWord into the PHP.. say: $uName = kriiv; and that is a correct username, it echoes out fine. So it just looks like the data is not coming across via the AJAX call.

Change (".login-username") to ("#login-username") and (".login-password") to ("#login-password") since you're using IDs and not classes. — Funk Forty Niner
– Funk Forty Niner, Commented Jul 26, 2014 at 5:22
Don't wait till your site gets hacked. Someone posted a question earlier about his getting hacked with similar code as yours. Your code is highly prone to SQL injection. Use mysqli_ with prepared statements, or PDO with prepared statements. — Funk Forty Niner
– Funk Forty Niner, Commented Jul 26, 2014 at 5:26
Plus, for password storage use CRYPT_BLOWFISH or PHP 5.5's password_hash() function. For PHP < 5.5 use the password_hash() compatibility pack. MD5 is old and considered broken. — Funk Forty Niner
– Funk Forty Niner, Commented Jul 26, 2014 at 5:26

웃웃웃웃웃 · Accepted Answer · 2014-07-26 05:43:26Z

1

Copy these it will work because login-username and login-password are id's not class.

username = $("#login-username").val();
password = $("#login-password").val();

edited Jul 26, 2014 at 5:43

웃웃웃웃웃

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answered Jul 26, 2014 at 5:22

Parimal

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Comments

Shah Rukh · Accepted Answer · 2014-07-26 05:22:57Z

0

Change this

   username = $(".login-username").val();
   password = $(".login-password").val();

to

   username = $("#login-username").val();
   password = $("#login-password").val();

. is used to get element having class # is used for having id , your element having id not class

answered Jul 26, 2014 at 5:22

Shah Rukh

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