1. Below is the PHP code I wrote that allows SQL injection with one parameter ie 'id' in this case.

  2. On trying SQL injection via putting this 2' ORDER BY 1;--' in the parameter (id field ), I get only SQL errors on the page.

  3. On executing the query directly on mysql server ie SELECT * FROM users WHERE id = '2' ORDER BY 1;--''; I get the below result that is 1 row set along with the error.

mysql> SELECT * FROM users WHERE id = '2' ORDER BY 1;--'';
| id | username | password  | creditcard       |
|  2 | John     | password! | 3123456769384659 |
1 row in set (0.00 sec)

ERROR 1064 (42000): 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 '--''' at line 1

PHP code


require_once 'login.php';
if (!isset($_GET['id']))
    echo <<<_END
    <pre>            <h1>WELCOME to the KINGDOM</h1>
    <form action = 'si2.php' method = 'GET'>
    id <input type = 'text' name = 'id'>

        <input type = 'submit' value = 'cl1ck M3'></pre></form>


    $id = stripslashes($_GET['id']);

$connection = mysql_connect($db_hostname,$db_username,$db_password);
if(!$connection) die ("Unable to connect with MySql " . mysql_error());

 mysql_select_db($db_database,$connection) or die('Could not connect with the database');

$query = "SELECT * FROM users WHERE id = '$id'";

$result = mysql_query($query);
    $rows = mysql_num_rows($result);
else { echo "Could not execute the Query:  <br>" . mysql_error();}

if($rows >= 1)

        for ($j=0 ; $j < $rows; ++$j)
                $row = mysql_fetch_row($result);
                echo "Hello $row[1]"."<br>";
                echo "Your Credit Card Number is  $row[3]"."<br><br>";
                echo $query."<br>";
                //echo $row[3];

    echo "<br><br><br>Sorry no rows/results could be fetched on query execution <br><br><br>";
    //echo $query;

My question is why my code isn't able to fetch the row set and displaying the table entries ? And in case I want to then what changes should I try.


2 Answers 2


Error and Problems in your injection

The problem is that you cannot execute multiple queries with mysql_query, and as you are injecting a ; you have multiple queries. The first one is SELECT * FROM users WHERE id = '2' ORDER BY 1 and the second one is --''.

The first one is the one that gives you the actual result in the command line, and the second one gives you your error (because it's not a valid query).

Note also that your injection: 2' ORDER BY 1;--' doesn't make all that much sense. Why do you have the ;? You don't really want to execute two queries, just one. And what is the last ' supposed to do? You already took care of the leftover ' by commenting it out via -- (at least in theory).


What you want to inject is 2' ORDER BY 1--, or, if that doesn't work, try using # (url encoded as %23 if you use a browser) instead of -- as comment character to cut of the rest of the query (the '), because it does not require an additional whitespace. In some cases you could also build the query in a way that the left over ' is used, such as OR '1' = '1 (this probably doesn't work for order by though).

  • Your clear description has explained where I's going wrong.
    – harveyD
    Commented Mar 26, 2015 at 1:18

There are possibly two issues with your attempt that cause it to fail:

  • Is that considered multiple queries? Commented Mar 25, 2015 at 18:07
  • @AbeMiessler I’m not sure.
    – Gumbo
    Commented Mar 25, 2015 at 18:08

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