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23,932 PointsSQL Injection Not Working?
For the fun of better understanding how SQL injections work, I decided to try an SQL injection of my own but it won't work.
I copied Alena's SQL injection exactly as is, but nothing happens? WHY?
details.php?id=2;DROP TABLE Media
I am using phpMyAdmin database via XAMPP and my project is being served on localhost. Not sure if this makes a difference?
1 Answer
Adriana Dodge
4,680 Pointsmistake when she adds media_id to the single_item_array function. the function should have Media.media_id on the select.
function single_item_array($id){
include("connection.php");
try{
$results = $db->query("SELECT Media.media_id, title, category, img, format, year,
publisher, isbn, genre
FROM Media
JOIN Genres ON Media.genre_id=Genres.genre_id
LEFT OUTER JOIN Books
ON Media.media_id = Books.media_id
WHERE Media.media_id = $id");
// echo"retreived Results";
}catch(Exception $e){
echo "Unable to retrieve results";
exhit;
}
$catalog = $results->fetch();
return $catalog;
}