Hello comrades.

I've got a dead end, how can I use SQL injection on this code? Your thoughts on this. I would be very grateful. )

<!DOCTYPE html> <html> <head> <meta charset="UTF-8"> <title></title> </head> <body> <h2>Для доступа к Бобруйской районной библиотеке введите Ваши учётные данные:</h2> <form method="get" action="?"> <p>Введите ваше имя</p> <input name="name" type="text"> <p>Введите ваш пароль</p> <input name="password" type="text"><br /> <input type="submit"> </form> <?php $mysqli = new mysqli("localhost", "root", "1111", "db_library"); if (mysqli_connect_errno()) { printf("Не удалось подключиться: %s\n", mysqli_connect_error()); exit(); } else { $mysqli->query("SET NAMES UTF8"); $mysqli->query("SET CHARACTER SET UTF8"); $mysqli->query("SET character_set_client = UTF8"); $mysqli->query("SET character_set_connection = UTF8"); $mysqli->query("SET character_set_results = UTF8"); } $name = filter_input(INPUT_GET, 'name'); $password = filter_input(INPUT_GET, 'password'); if ($result = $mysqli->query("SELECT * FROM `members` WHERE name = '$name' AND password = '$password'")) { while ($obj = $result->fetch_object()) { echo "<p><b>Ваше имя: </b> $obj->name</p> <p><b>Ваш статус:</b> $obj->status</p> <p><b>Доступные для Вас книги:</b> $obj->books</p><hr />"; } } else { printf("Ошибка: %s\n", $mysqli->error); } $mysqli->close(); ?> </body> </html>

    1 answer 1

    The first thing you need to do: send data containing a quote in the input path. For example, the password 'OR 1--

    "-" in MySQL is a comment, that is, we comment on all subsequent actions, and our query takes the form:

     SELECT * FROM `members` WHERE name = '$name' AND password = 'пароль' OR 1'"

    In this case, we will get the data of all users from this table.