我在PHP中新,我试图做在PHP 5的工作,所以我有称为参数化的功能,但它给了我异常“警告:缺少参数1”这里是我的课PHP的警告:缺少参数1
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();
}
}
,这是我怎么称呼它
include_once('include/classes/edituser.php');
$objPage = new EditUser();
$objPage->editUser($_GET['id']);
但它显示我的警告,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ŧ请问我有什么不对,因为我认为我正在使用正确的方式来称呼它。
so'$ _GET ['id']'没有设置。永远不要相信用户的输入! – k102
如果你使用'$ _GET',那么你应该在你的查询字符串上有一个id:'http:// localhost?id = 1',其中id是1 –
我对我的查询字符串 – user2930808