php db query to display certain user data (Smarty included) - php

Question

I have managed to create a login form which GETS user data from database and displays user data, but i would now like to display the table_name to which that user is related to.
The thing is i will have to use an JOIN because it needs to query the data from 2 tables.
My Problem
I keep getting an error, im not sure if my query in my php file has the wrong syntax and if my method is just the wrong approach.
Connection
Keep in mind my database connections ($db) works perfectly as the user can already log in with username and password
Error
Notice: Undefined variable: rows in C:\xampp\htdocs\score4score\table.php on line 42
Database: tables
table: sc_tables
+------+----------+-------------+
| t_id | user_id | table_name |
+------+----------+-------------+
| 1 | 1 | bobs |
+------+----------+-------------+
table: sc_scores
+------+----------+-------------+-------+
| s_id | user_id | t_id | score |
+------+----------+-------------+-------+
| 1 | 1 | 1 | 50 |
+------+----------+-------------+-------+
PHP
<?php
include("config.php");
session_start();
$userId = (isset($_GET["user_id"]) && is_numeric($_GET["user_id"]) && (int)$_GET["user_id"] > 0) ? (int)$_GET["user_id"] : 0;
$query = "SELECT `a`.`table_name` FROM `sc_tables` AS `a` JOIN `sc_scores` AS `b` ON `a`.`t_id` = `b`.`t_id`";
$result = mysqli_query($db, $query);
$rows = array();
while ($row = mysql_fetch_array($result)) {
$rows[] = $row;
}
$smarty->assign('rows', $rows);
include("header.php");
$smarty->display('records.tpl');
?>
TPL (records.tpl)
<div>
<div>
The current rankings table:
<table>
{foreach from=$rows item="row"}
<tr>
<td>{$row.table_name}</td>
</tr>
{/foreach}
</table>
</div>
</div>
It would be much appreciated if someone can just point me in the right direction. thanks

Answer 1

There is no table_id field in the tables. So, you should have:
$query = "SELECT `a`.`table_name` FROM `sc_tables` AS `a` JOIN `sc_scores` AS `b` ON `a`.`t_id` = `b`.`t_id`";
The error says that the value FALSE is passed as argument to mysql_fetch_array($result). E.g. your mysqli_query() returns FALSE, not a mysqli_result object, as it should.
If problem persists after upper changes, without seeing any error, then you should use also mysqli_errno(), mysqli_error()and mysqli_error_list().
Good luck!