两次使用的PHP类参数

Question

我是新的OOP，我正在切换我所有的网站代码到它！我目前正在编写一个获取用户信息的类，并将最终更新它。

我使用的代码如下：

<?php

    require("C:\wamp\www\postin'\db_connection.php");
    session_start();

    class user {

        public function __construct($userid, $connection, $information) {
            $this->userid = $userid;
            $this->connection = $connection;
            $this->information = $information;
        }   

        public function user_information($userid, $connection, $information) {
            $query = "SELECT * FROM users WHERE id = :id";
            $params = array(':id' => $userid);
            try{
                $stmt = $connection->prepare($query);
                $result = $stmt->execute($params);
            }                               
            catch(PDOException $ex){
                echo ("Failed to run query: " . $ex->getMessage());
            }                       
            $columns = $stmt->fetch();
            return $columns["$information"];
        }

    }

    $username = new user($_SESSION["logged_in"], $connection, "username");
    echo $username->user_information($_SESSION["logged_in"], $connection, "username");
?>

现在，正如您在最后两行代码中所看到的(最后一行代码)，我必须使用参数两次。基本上，第一个参数表示ID是什么，第二个参数表示什么是$connection，第三个参数是我想从数据库中获取的内容。那我做错什么了？我是不是定义了一些我不需要的东西？

编辑

以下内容是否也有效？

<?php

require("C:\wamp\www\postin'\db_connection.php");
session_start();

class user {

    public function user_information($userid, $connection, $information) {
        $query = "SELECT * FROM users WHERE id = :id";
        $params = array(':id' => $userid);
        try{
            $stmt = $connection->prepare($query);
            $result = $stmt->execute($params);
        }                               
        catch(PDOException $ex){
            echo ("Failed to run query: " . $ex->getMessage());
        }                       
        $columns = $stmt->fetch();
        return $columns["$information"];
    }

}

$username = new user();
echo $username->user_information($_SESSION["logged_in"], $connection, "username");
?>

就像这是正确的，还是错的.？

Answer 1

如果user类拥有它作为数据成员所需的所有信息，那么user_information就不需要接受任何参数：

public function user_information() {
    $query = "SELECT * FROM users WHERE id = :id";
    $params = array(':id' => $this->userid);
    try{
        $stmt = $this->connection->prepare($query);
        $result = $stmt->execute($params);
    }                               
    catch(PDOException $ex){
        echo ("Failed to run query: " . $ex->getMessage());
    }                       
    $columns = $stmt->fetch();
    return $columns[$this->information];
}

Answer 2

由于你有很多关于一个课程的工作方式和OOP的问题，我会尝试给你一个小的方向。没有标准的方法来建设你的班级。你是决定什么去向的人，就属于这个类的东西和需要注入的东西而言。这只是告诉你，你不能固步自封。你需要对它有一个感觉，并建立一个逻辑。我选了你的课，并重新编写了一些评论。希望这会对你有所帮助。祝好运!

    <?php

require ("C:\wamp\www\postin'\db_connection.php");
session_start();

class user {

    public $dbconnection;

    public function __construct($connection) {
        /**
         * Your user class interacts with the database.
         * Inject the connection here and set your
         * global class variable.
         */
        $this -> dbconnection = $connection;
    }

    public function user_information($userid, $column) {
        /**
         * The userid and column are specific for this
         * method action. No need to set these variables
         * in the global scope of the class.
         */

        $query = "SELECT" . $column . " FROM users WHERE id = :id";

        $params = array(':id' => $userid);
        try {
            $stmt = $this -> dbconnection -> prepare($query);
            $stmt -> execute($params);
        } catch(PDOException $ex) {
            echo("Failed to run query: " . $ex -> getMessage());
        }
        $result = $stmt -> fetch();

        return $result;
    }

}

$username = new user($connection);
echo $username -> user_information($_SESSION["logged_in"], $information);
?>