I am trying to create a login page to demonstrate a smple sql injection attack. Below is the php code that receives input from the login.html:
<?php
define('DB_HOST', 'localhost');
define('DB_NAME', 'sql-injection');
define('DB_USER','root');
define('DB_PASSWORD','');
$con=@mysql_connect(DB_HOST,DB_USER,DB_PASSWORD) or die("Failed to connect to MySQL: " . mysql_error());
$db=mysql_select_db(DB_NAME,$con) or die("Failed to connect to MySQL: " . mysql_error());
/* $ID = $_POST['user']; $Password = $_POST['pass']; */
function SignIn()
{
session_start(); //starting the session for user profile page
if(!empty($_POST['user']) and !empty($_POST['pass'])) //checking the username and password which is coming from Sign-In.html, are they empty or have some text
{
$query = mysql_query("SELECT * FROM UserName where userName = '$_POST[user]'") or die(mysql_error());
$row = mysql_fetch_array($query) or die(mysql_error());
if(!empty($row['userName']) AND !empty($row['pass']))
{
if ($_POST['pass'] == $row['pass'])
{
echo "SUCCESSFULLY LOGIN TO USER PROFILE PAGE...";
}
else
{
echo "SORRY... YOU ENTERD WRONG ID AND PASSWORD... PLEASE RETRY...";
}
}
else
{
echo "The db is corrupt";
}
}
else{
echo "The username and password fields cant be empty";
}
}
if(isset($_POST['submit']))
{
SignIn();
}
?>
The db has one table username with columns
UsernameID userName pass
1 mrboolnew mrbool123
I am trying to enter this in the username column:
Y';UPDATE username SET userName='mrboolnew1' WHERE userName = 'mrboolnew';
but I get the below error:
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'UPDATE username SET userName='mrboolnew1' WHERE userName = 'mrboolnew';'' at line 1
Can someone help me out with the injection query. Thanks in advance.
You have added protection for protecting against injection attacks, perhaps without even knowing in. mysql_query()
sends only a single query to the database (see here). Semicolons are not allowed; nor are multiple queries.
You'll have to try harder to inject your own code.