Question

我写了这段代码，但它没有用。它基本上没有做任何事情。此表单放在表单所在的login.php上。表单操作设置为php self，因此它将信息发送给自己，我正在尝试使用此代码检查信息是否有效;

if ($_POST['username'] != "" && $_POST['password'] != "") {
$data = mysqli_query($dbcon, "  SELECT * 
                                FROM  `users` 
                                WHERE  `user_name` LIKE  '".$_POST['username']."'
                                AND  `user_password` LIKE  '".$_POST['password']."'") or die(mysql_error());

        // if this user exists
        if ($data->num_rows == 1) {
                $result_row = $data->fetch_object();
                $_SESSION['user_name'] = $result_row->user_name;
                $_SESSION['user_password'] = $result_row->user_password;
                $_SESSION['user_logged_in'] = 1;
        } else {
            "Wrong Password";
        }

}

Answer 1

解决了其他人的帮助问题！

谢谢大家！

if ($_POST['username'] != "" && $_POST['password'] != "") {
$data = mysqli_query($dbcon, "  SELECT * 
                                FROM  `users` 
                                WHERE  `user_name` =  '".$_POST['username']."'
                                AND  `user_password` =  '".$_POST['password']."'") or die(mysqli_error());

        // if this user exists
        if ($data->num_rows == 1) {
                session_start();
                $result_row = $data->fetch_object();
                $_SESSION['user_name'] = $result_row->user_name;
                $_SESSION['user_password'] = $result_row->user_password;
                $_SESSION['user_logged_in'] = 1;
        } else {
            echo "Wrong Password";
        }

}

PHP连接数据库以检查用户登录信息

1 个答案: