php - 如何将登录成员(member)下的成员(member)表与玩家表合并？

Question

我正在制作一个网站，您需要在其中注册，然后创建一个角色来玩。我如何将注册页面中使用的表格与玩家的表格结合起来，以便玩家始终获得他创建的角色。

我有一个表members，用于存储注册用户以及角色的 table 玩家

我查看了 session_id，但据我了解，我无法将数据存储到 mysql 表中。

我还考虑过在创建时将用户名和电子邮件从成员表添加到玩家表中，但我失败了。我想我可以说，如果两个表中的用户名和 ID 相同，则该玩家将被使用。

我对这一切都很陌生。

这是注册码

<?php
include_once 'db_connect.php';
include_once 'psl-config.php';

$error_msg = "";

$now = time();


if (isset($_POST['username'], $_POST['email'], $_POST['p'])) {
    // Sanitize and validate the data passed in
    $username = filter_input(INPUT_POST, 'username', FILTER_SANITIZE_STRING);
    $email = filter_input(INPUT_POST, 'email', FILTER_SANITIZE_EMAIL);
    $email = filter_var($email, FILTER_VALIDATE_EMAIL);
    if (!filter_var($email, FILTER_VALIDATE_EMAIL)) {
        // Not a valid email
        $error_msg .= '<p class="error">The email address you entered is not valid</p>';
    }

    $password = filter_input(INPUT_POST, 'p', FILTER_SANITIZE_STRING);
    if (strlen($password) != 128) {
        // The hashed pwd should be 128 characters long.
        // If it's not, something really odd has happened
        $error_msg .= '<p class="error">Invalid password configuration.</p>';
    }

    // Username validity and password validity have been checked client side.
    // This should should be adequate as nobody gains any advantage from
    // breaking these rules.
    //

    $prep_stmt = "SELECT id FROM members WHERE email = ? LIMIT 1";
    $stmt = $mysqli->prepare($prep_stmt);

   // check existing email  
    if ($stmt) {
        $stmt->bind_param('s', $email);
        $stmt->execute();
        $stmt->store_result();

        if ($stmt->num_rows == 1) {
            // A user with this email address already exists
            $error_msg .= '<p class="error">A user with this email address already exists.</p>';
                        $stmt->close();
        }
                $stmt->close();
    } else {
        $error_msg .= '<p class="error">Database error Line 39</p>';
                $stmt->close();
    }

    // check existing username
    $prep_stmt = "SELECT id FROM members WHERE username = ? LIMIT 1";
    $stmt = $mysqli->prepare($prep_stmt);

    if ($stmt) {
        $stmt->bind_param('s', $username);
        $stmt->execute();
        $stmt->store_result();

                if ($stmt->num_rows == 1) {
                        // A user with this username already exists
                        $error_msg .= '<p class="error">A user with this username already exists</p>';
                        $stmt->close();
                }
                $stmt->close();
        } else {
                $error_msg .= '<p class="error">Database error line 55</p>';
                $stmt->close();
        }

    // TODO: 
    // We'll also have to account for the situation where the user doesn't have
    // rights to do registration, by checking what type of user is attempting to
    // perform the operation.

    if (empty($error_msg)) {
        // Create a random salt
        //$random_salt = hash('sha512', uniqid(openssl_random_pseudo_bytes(16), TRUE)); // Did not work
        $random_salt = hash('sha512', uniqid(mt_rand(1, mt_getrandmax()), true));

        // Create salted password 
        $password = hash('sha512', $password . $random_salt);


        // Insert the new user into the database
        // Add here wat you want to add into the database at account creation  

        if ($insert_stmt = $mysqli->prepare("INSERT INTO members (username, email, password, salt, accdate) VALUES (?, ?, ?, ?, now())")) {
            $insert_stmt->bind_param('ssss', $username, $email, $password, $random_salt);

            // Execute the prepared query.
            if (! $insert_stmt->execute()) {
                header('Location: ../error.php?err=Registration failure: INSERT');
            }
        }
        header('Location: ./register_success.php');
    }
}

有人能指出我正确的方向吗？谢谢

Answer 1

需要阅读的内容有很多，一般来说，您可能希望将用户表上的用户 ID 与角色相关联。 ( user_id 将是指向 user 表上的 id 的字符的外键。两个表都有 id 的主键，它应该自动递增)，然后您必须维护关系，以便当他们创建字符时您可以插入当前的字符将 user_id 登录到表中，如果您想将它们限制为一个字符，您可以使该列 ( user_id ) 唯一。

例如...

table user, 
id, name, login, password  etc...

table character
id, user_id, character_name, hp, etc..

然后你可以像这样加入表格

SELECT u.*, c.* FROM users AS u JOIN characters AS c ON u.id = c.user_id WHERE u.id = $user_id.

这将(理论上)为您提供用户表和字符表中 $user_id 的所有记录(假设它们在两个表中都有记录，如果它们没有或可能没有字符记录，则使用 LEFT JOIN )

这里有一个关于一些基本数据库关系的很好的教程。

http://code.tutsplus.com/articles/sql-for-beginners-part-3-database-relationships--net-8561