如何正确地从构造函数中提取“type”mysqli

Question

我遇到了一个问题，我试图从数据库连接类中获取用于查询的数据库链接，但分配显示它是一个对象而不是我需要运行查询的mysqli链接。我收到错误：“PHP警告：mysqli_query（）期望参数1是mysqli，给定的对象...”。

这里是我的代码为我的DB连接类：

class DB_Connect { 

function __construct() { 

} 

function __destruct() { 

} 
public function connect() { 

    $con = mysqli_connect('localhost', 'user', 'password','database') or die("Error " . mysqli_error($con)); 
//var_dump($con); 
if (!$con) { 
    throw new Exception('Could not connect to database server'); 
} else { 
    return $con; 
} 
} 

这里是我的代码为DB功能：

class DBfns { 

private $db; 

function __construct() { 
    require_once 'DB_Connect.php'; 
    // connecting to database 
    $this->db = new DB_Connect(); 
    $this->db->connect(); //Needs to be changed to mysqli object 
    var_dump($this->db); //Currently showing as "object(DB_Connect)#2 (0) {}" 

} 

// destructor 
function __destruct() { 

} 
public function getUser($uname) { 
$res = mysqli_query($this->db, "SELECT user_id FROM users WHERE uname = '$uname' "); 
//var_dump($this->db); 

} 
} 

预先感谢您的帮助！

Answer 1

你需要做更多这样的：

class DB_Connect { 

    private $connection; 

    public function __construct() { 

     $this->connection = mysqli_connect('localhost', 'user', 'password','database') or die("Error " . mysqli_error($con)); 

     if (!$this->connection) { 
      throw new Exception('Could not connect to database server'); 
     } 
    } 

    public function getConnection() { 
     return $this->connection; 
    } 
} 

然后在DBfns类

$this->db = new DB_Connect(); 

后来

$res = mysqli_query($this->db->getConnection(), "sql");