Question

我正在尝试在登录脚本上创建强力保护。虽然应该更新帐户锁定的数据库的查询将无法正常工作。我开始感到沮丧，因为我在过去的半小时里一直在看这个代码..

无论如何这里是代码：

if(isset($_POST['submit'])) {
$mysqli = getConnected('*','*','*','*');
$username = $mysqli->escape_string($_POST['username']);
$password = $mysqli->escape_string($_POST['password']);
$hash = sha1($password);
$query = "SELECT * FROM users WHERE username='$username'";
$result = $mysqli->query($query);
$data = $result->fetch_assoc();

    if($data['lock'] == 0) {
        if($hash == $data['password']) {
        $_SESSION['loggedin'] = true;
        $_SESSION['username'] = $username;
        $result->free();
        $mysqli->close();
        header("Location: index.php?p=start");
        } else {
            if($data['login_attempts'] != 0) {
                $attempts_left = $data['login_attempts'] - 1;
                $query2 = "UPDATE users SET login_attempts='$attempts_left' WHERE username='$username'";
                $mysqli->query($query2);
                echo "Inloggningen misslyckades, du har ". $attempts_left ." försök kvar.";
            } else {
                $query3 = "UPDATE users SET lock='1' WHERE username='$username'";
                $mysqli->query($query3);
                echo "Ditt konto har låsts.";
            }
        }
    } else {
        echo "Ditt konto är låst, kontakta webmaster för att återställa det.";
    }
}

问题在于执行$ query3，尽管它根据var_dump正确执行。

我需要一些第二眼，以确保我不会发疯！

Answer 1

问题是lock是reserved word in mysql。您需要将查询更改为：

$query3 = "UPDATE users SET `lock`='1' WHERE username='$username'";
                            ^^^^^^ use back-ticks to escape reserved words in mysql

PHP更新查询不会更新

1 个答案: