Ошибка - Warning: mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, bool given in C:\xam

Вот код:

PHP:

1. $ghost = "SELECT * FROM matches ";
2. $result = mysqli_query($db,$ghost);
3. echo '<table border = "1" >'  . '<td>' . 'Команда 1'. '</td>' . '<td>' . 'Команда 2'. '</td>' . '<td>' . 'Дата матча'. '</td>'. '<td>' . 'Стадион'. '</td>'.'<td>'. ' Цена ' . '</td>'.'<td>'. ' Кол-во оставшихся мест ' . '</td>'.'<td>'. ' Купить ' . '</td>' .  '</tr>';
4. for ($i = 0; $i<mysqli_num_rows($result); $i++) {
5.  
6.     $result_arr = mysqli_fetch_assoc($result);
7.     $none = $result_arr['team_id'];
8.     $teamone = "SELECT name FROM team where id = '$none' ";
9.     $teamres = mysqli_query($db,$teamone);
10.     $teamone_arr = mysqli_fetch_assoc($teamres);
11.     $nane = $result_arr['team2_id'];
12.     $teamtwo = "SELECT name FROM team where id = '$nane' ";
13.     $teamt = mysqli_query($db,$teamtwo);
14.     $teamtwo_arr = mysqli_fetch_assoc($teamt);
15.     $st = $result_arr['stadium_id'];
16.     $stadium = "SELECT name FROM stadium where id = '$st' ";
17.     $stad = mysqli_query($db,$stadium);
18.     $stadium_arr = mysqli_fetch_assoc($stad);
19.     $data = $result_arr['date'];
20.     $res = "SELECT COUNT (*) FROM ticket WHERE date =  '$data'";
21.     $minus = mysqli_query( $db,$res );
22.     echo $minus;
23.     $minus_arr = mysqli_fetch_assoc($minus);
24.     $sold_tickets = $minus_arr['count'];
25.     $plus = $result_arr['place'];
26.     $place = $plus - $sold_tickets;
27.     echo   '<td>'  . $teamone_arr['name'] . '</td>' . '<td>'  . $teamtwo_arr['name'] . '</td>' . '<td>' . $result_arr['date'] . '</td>' . '<td>'  . $stadium_arr['name']  . '</td>' . '<td>'  . $result_arr['cost'] . '</td>'. '<td>'  .  $place . '</td>'. '<td>'  .  '<form> <input type="text" name="num" placeholder="Введите кол-во билетов"><br><input type= "submit" value="Купить"></form>' . '</td>'. '</tr>' ;
28. }

Не понимаю что не так

От модератора: Используйте кнопку </> для вставки кода