Question

我构建了一个User类，它试图在MySQL数据库中查找用户的ID。如果找到它，它会将变量SQL_ID设置为此值，否则将其保留为空。

一个单独的方法（IsValid），稍后调用，返回一个布尔值，告诉我用户是否确实存在。

我很好奇是否有人想在这里评论我的设计，并且可能提供更优雅的解决方案。我承认PHP不是我的主要语言，在用非静态类型的语言花费太多时间后，我可能会感到有些OCDish。也许我正在寻求验证这种设计是否合理。

// User -> class for passing around user information. Should only pass around the UserID (a unqiue SQL ID), for security reasons, in a Session object.
    class User {
        private $SQL_ID = "";
        //@todo: Get the User object to actually talk to the other classes. Lol.

        public function __construct($Username, $Password) {
            // Probably want to Base64 encode the values going into and out of the MySQL database, to prevent a SQL Injection attack.
            $query = "SELECT [UserID] FROM [Users] WHERE [Username] = '" . base64_encode($Username) . "' AND [Password] = '" . base64_encode($Password) . "';";
            $data = SQL::DataQuery($query);

            $this->SQL_ID = $data["UserID"];

        }

        // Boolean function to tell us if we have a valid user. Might be able to merge this into the constructor.
        public function IsValid() {
            if($this->SQL_ID == "") {
                return false;
            }

            return true;
        }

        public function GetUserID() {
            return $this->SQL_ID;

        }

        // private $Query = "SELECT [UserID] FROM [Users] WHERE [Username] = '' AND PASSWORD = '';"; // Prototype User query (for selecting a UserID).


    }

Answer 1

除非登录成功，否则我认为您不应该拥有User对象。我会使用另一个类（类似于Authenticator）来负责检查数据库的用户名和密码，并仅在找到有效用户时返回User对象。

设计更优雅的User对象

1 个答案: