PHP警告: Missing argument 1

Question

我是 php 新手，我正在尝试在 php 5 中工作，所以我调用了一个参数化函数，但它给了我异常“警告:缺少参数 1”，这是我的类

class EditUser extends DBConn
{

private $id;
function editUser($id)
{
    $this->id = $id;
    echo $id;
    die;

    $rows =array();
    self::Set_DBConni();
    $mysqli = self::get_Conn();
    $result = $mysqli->query("SELECT * FROM users where id ='".$id."' ");

        while($row = $result->fetch_row())
        {
            $rows[] = $row;

        }


        return $rows;

        /* free result set */
        $result->close();

        /* close connection */
        $mysqli->close();

}


 }

这就是我所说的

include_once('include/classes/edituser.php'); 

$objPage = new EditUser();

$objPage->editUser($_GET['id']);

但它向我显示了警告，那就是

Warning: Missing argument 1 for EditUser::editUser(), called in E:\xampp\htdocs\WaleedWork\claremont\admin\edit_user.php on line 45 and defined in E:\xampp\htdocs\WaleedWork\claremont\admin\include\classes\edituser.php on line 8

Notice: Undefined variable: id in E:\xampp\htdocs\WaleedWork\claremont\admin\include\classes\edituser.php on line 10

Notice: Undefined variable: id in E:\xampp\htdocs\WaleedWork\claremont\admin\include\classes\edituser.php on line 11

请告诉我出了什么问题，因为我认为我使用了正确的方式来调用它。

Answer 1

你的editUser方法是一个构造函数，你可以这样做

$objPage = new EditUser($_GET['id']);