Ето и кода
Дава ми това.
Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, string given in /home/a7816969/public_html/admin/koki.php on line 8
Как да оправя проблема?
PHP:
<?php
include("conf.php");
if ($_COOKIE['id'] && $_COOKIE['pass']){
$idCookie = filter_input(INPUT_COOKIE,'id');
$passCookie = filter_input(INPUT_COOKIE,'pass');
$logged = ("SELECT * from users WHERE id='$idCookie' AND password = '$passCookie'");
$result = mysqli_query($conn, $logged);
$user = mysqli_fetch_array($logged);
}
?>
Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, string given in /home/a7816969/public_html/admin/koki.php on line 8
Как да оправя проблема?