Здравствуйте. Подскажите пожалуйста как исправить ошибку. При запуске сайта на главной странице сайта выводится текст- Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/u489318331/public_html/blocks/footer.php on line 14,16,18,20,22,24,26,28,30,32,34,36,38,40. Вот линии скрипта с ошибками -</div>
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</td>
<td class="secondcol"><div id="secondcol">
<div class="accent">
<div class="accent-left"></div>
<div class="accent-right"></div>
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<?
require ("connectdb.inc.php");
$sql0=mysql_query("select*from sait");
$num_sait=mysql_fetch_array($sql0);
$sql=mysql_query("select*from users");
$num=mysql_num_rows($sql);
$sql1=mysql_query("select*from orders");
$num_nord=mysql_num_rows($sql1);
$sql2=mysql_query("select*from orders2");
$old_ord=mysql_num_rows($sql2);
$sql3 = mysql_query("SELECT * FROM `users` where rang='0'");
$ur0=mysql_num_rows($sql3);
$sql4 = mysql_query("SELECT * FROM `users` where rang='1'");
$ur1=mysql_num_rows($sql4);
$sql5 = mysql_query("SELECT * FROM `users` where rang='2'");
$ur2=mysql_num_rows($sql5);
$sql6 = mysql_query("SELECT * FROM `users` where rang='3'");
$ur3=mysql_num_rows($sql6);
$sql7 = mysql_query("SELECT * FROM `users` where rang='4'");
$ur4=mysql_num_rows($sql7);
$sql8 = mysql_query("SELECT * FROM `users` where rang='5'");
$ur5=mysql_num_rows($sql8);
$sql9 = mysql_query("SELECT * FROM `users` where rang='6'");
$ur6=mysql_num_rows($sql9);
$sql10 = mysql_query("SELECT * FROM `users` where rang='7'");
$ur7=mysql_num_rows($sql10);
$sql11 = mysql_query("SELECT * FROM `users` where rang='8'");
$ur8=mysql_num_rows($sql11);
$sql12=mysql_query("select*from users where date between subdate(now(), interval 0 day) and now()");
$row00=mysql_num_rows($sql12);
?>