Question

我试图继承所有的方法和属性从超全球的Mesqli 类到我的DB类。

这是DB类:

class DB extends mysqli
{
    protected $mysqli;

    public function __construct () {
        // connect to MySQL
        $mysqli = new mysqli( host ,  username ,  password ,  dbname );

        // output error if unable to connect
        if ($mysqli->connect_errno) {
            echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " .     $mysqli->connect_error;
            exit;
        }
    }
}

以下是主要的会议课:

class Meetings
{

    function __construct () {
        require_once( ../include/classes/db.class.php );
        $db = new DB();

        $field = $db->real_escape_string($_POST[ about ]);

    }

}

在会议课上,我想通过即时DB课来称呼真实的越野字符串方法由超级全球的Mesqli课继承

我得到这个错误 : 警告: Mysqli:: real_ escape_string () [mysqli. real- escape- string]: 无法获取 DB

Answer 1

Oops! You created the DB object, and the db object creates an another instance of the mysqli class. You should not call new mysqli, but you should call

parent::__construct( host ,  username ,  passw或d ,  dbname )

或

    parent::mysqli( host ,  username ,  passw或d ,  dbname )

I don t know which is the good f或mat f或 the mysqli construct或.

在此之后, 您不需要 DB 类中的 Mysql 属性, 因为在这种情况下, 您的 Mysqli 属性等于 DB 类的美元。所以, 不是 $mysqli- gt; connect_ errno, 而是$this- gt; connect_ errno 是正确的。

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