PHP - 使用類獲取用戶信息

Question

假設與數據庫的連接以及對錶和單元格的所有引用都是正確的......我怎麼能得到這樣的工作？

class User 
{ 

    private $_display; 
    private $_email; 

    public function __construct($username) 
    { 

     $fetch_user = mysql_query("SELECT * FROM `registered_users` WHERE `user_name`='$username'"); 
     $fetch_user = mysql_fetch_array($fetch_user); 

     $this->_display = $fetch_user['user_display']; 
     $this->_email = $fetch_user['user_email']; 
    } 

} 
$person1 = new User('username'); 
echo "Information: " . print_r($person1, TRUE); 

問題是它返回什麼都沒有。調試時不會拋出錯誤或任何東西。它是可行的方法嗎？ ：S

Answer 1

這裏是大致我會做什麼：

<?php 

    class User{ 

    private $username; 
    private $data; 

    public function __construct($username){ 
     $this->username = $username; 
     if($this->valid_username()){ 
     $this->load(); 
     } 
    } 

    private function load(){ 
     // Let's pretend you have a global $db object. 
     global $db; 
     $this->data = $db->query('SELECT * FROM registered_users WHERE user_name=:username', array(':username'=>$this->username))->execute()->fetchAll(); 
    } 

    public function save(){ 
     // Save $this->data here. 
    } 


    /** 
    * PHP Magic Getter 
    */ 
    public function __get($var){ 
     return $this->data[$var]; 
    } 

    /** 
    * PHP Magic Setter 
    */ 
    public function __set($var, $val){ 
     $this->data[$var] = $val; 
    } 


    private function valid_username(){ 
     //check $this->username for validness. 
    } 

    // This lets you use the object as a string. 
    public function __toString(){ 
     return $this->data['user_name']; 
    } 

    } 

如何使用：

<?php 

    $user = new User('donutdan'); 
    echo $user->name; //will echo 'dan' 
    $user->name = 'bob'; 
    $user->save(); // will save 'bob' to the database