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我想将mysqli对象传递给类。我在一个文件中创建对象:包含在一个页面将mysqli对象传递给类
$host = 'host';
$user = 'username';
$pw = 'pw';
$db = 'db';
$link = new mysqli($host,$user,$pw,$db);
:
include 'php/mysql.php';
include 'php/classes/Class.Dataconnector.php';
$dc = new Dataconnector($link);
和类看起来是这样的:
class Dataconnector {
protected $_link;
protected $_stub;
function __construct(mysqli $link) {
$_link = $link;
}
public function getPageContent($stub) {
$query = "select * from contents where pageId = (select id from pages where stub = '$stub')";
$result = mysqli_query($_link,$query);
return $result;
}
}
但我得到这个错误:
Warning: mysqli_query() expects parameter 1 to be mysqli, null given in \php\classes\Class.Dataconnector.php on line 18
出了什么问题?
它不是全局的,它是一个对象属性。 –
@MadaraUchiha对不起,很好的术语(不知道我在想什么= P)。更新我的措辞=] – newfurniturey