Ошибка Trying to access array offset on value of type null

Выдает ошибку "Warning: Trying to access array offset on value of type null in D:\wamp64\www\salarygive.php on line 20".

PHP:

1. <?php
2. $host = "127.0.0.1";
3. $name = "root";
4. $pass = "";
5. $con = mysqli_connect($host, $name, $pass, "mysql");
6. $sql = "SELECT * from gamers";
7.  
8. if ($result = mysqli_query($con, $sql)) {
9.  
10.     // Return the number of rows in result set
11.     $count = mysqli_num_rows( $result );
12. }
13. for ($count; $count >= 0; $count--) {
14.     $zapros = "SELECT * FROM `gamers` WHERE \"" . $count ."\"";
15.     //echo $zapros;
16.     $sql = mysqli_query($con, $zapros);
17.     $res  = mysqli_fetch_array($sql);
18.     //$u = $res["money"] + $res["salary"];
19.     $zapros = "UPDATE `gamers` SET `money`= \"" . $res["money"] + $res["salary"] . "\" WHERE \"" . $count ."\"";
20.     $sql = mysqli_query($con, $zapros);
21. }
22. ?>
23. <form action="indexadmin.php">
24.     <p>Успешно</p>
25.     <input type="submit" value="Домой" />
26. </form>

Благодарю за помощь заранее. Ругается на строку, которая начинается на $zapros = "UPDATE `gamers` SET `money`