從MySQL數據庫獲取結果時出錯

Question

這是我的類，用於連接到數據庫，但是當我創建該類的實例時，它給了我這個錯誤：Fatal error: Call to undefined method DB::query()在C：\ Apache24 \ htdocs \ core \ DB.class.php上第21行

<?php 

class DB { 
    protected $db_name = "data_db"; 
    protected $db_user = "root"; 
    protected $db_pass = "root"; 
    protected $db_host = "localhost"; 

    public function __construct() { 
     $db = new mysqli($this->db_host, $this->db_user, $this->db_pass, $this->db_name); 
     if ($db->connect_errno) { 
      printf("Connection failed: %s\n", $db->connect_error); 
      exit(); 
     } else 
      return $db; 
    } 
} 

$db = new DB(); 

$data_row = $db->query("SELECT * FROM `users`"); 
while($data = $data_row->fetch_assoc()) 
    echo $data["first_name"] . '<br>'; 

?> 

致命錯誤：調用未定義的方法DB ::查詢（）在C：\ Apache24 \ htdocs中\核心\上線DB.class.php 21

Answer 1

構造函數沒有返回自定義值，他們返回類的實例。要測試這個使用var_dump($db);和輸出將是object(DB) ...。將連接保存爲對象的屬性，如：

public function __construct() { 
    $this->mysqli = new mysqli($this->db_host, $this->db_user, $this->db_pass, $this->db_name); 
} 

$db = new DB(); 
$data_row = $db->mysqli->query("SELECT * FROM `users`");