[SOLVED] Is my SQL request secure with mysql_real_escape_string?

Issue

Below is a page that handles a login script and I am wondering if I have put it any security holes. I have been reading articles on protecting from injections and others and wanted to make sure that my code is secure.

It is submitted via ajax and returns JSON based on the login being correct or not.

<?php
ob_start();
session_start();
include ("config.inc.php");
include ("jsonEncode.php");

// ausername and apassword sent from form
$ausername = '';
$apassword = '';
$ausername = mysql_real_escape_string(stripslashes($_GET['username']));
$apassword = mysql_real_escape_string(stripslashes($_GET['password']));

$sql    = "SELECT * FROM admin WHERE ausername='$ausername' AND apassword='$apassword' LIMIT 1";
$result = mysql_query($sql) or die(mysql_error());

$data   = mysql_fetch_array($result);
$count  = mysql_num_rows($result);

if($count==1){
    $_SESSION['ausername'] = $ausername;
    $_SESSION['apassword'] = $apassword;
    $_SESSION['admin_id']  = $data['a_id'];
    $a_id = $data['a_id'];
    $_SESSION['LastLogin'] = $data['last_login'];
    $query = "UPDATE admin SET last_login = Now() WHERE `a_id`= $a_id";
    mysql_query($query);
    //echo $query;
    $_SESSION['aloggedin'] = "1234";
    // valid
    $var = array('avalid' => 1, 'ausername' => $ausername, 'apassword' => $apassword);
    print php_json_encode($var);
}else{
    // invalid
    $var = array('avalid' => 0, 'ausername' => $ausername, 'apassword' => $apassword);
    print php_json_encode($var);
}
?>

Solution

You might want to use the POST method rather than GET with the login form, otherwise their password will appear in the URL and URLs aren’t very secure (they might get bookmarked or sent to another server as a referral URL, for example).

Answered By – Paige Ruten

Answer Checked By – Katrina (BugsFixing Volunteer)