Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource

arkadaslar daha onceden burda bı hata yokken sımdı bana "Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in ..footer.php on line 16" hatası veriyor yardımcı olursanız sevinirim kac gundur ugrasıyorum bı sonuc cıkaramadım

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			<li id="categories">

				<h2>Ýstatistik</h2>

				<ul>
(işte hata veren satır,garip kısmı 16.satır burası ve kod yok burda)
			<?=$ist?>

				</ul>

			</li>



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				<h2>Etiketler</h2>

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<? $etiket=mysql_query("select * from tag order by rand() limit 50"); 

   while ($eti=mysql_fetch_array($etiket))  { ?>





							<a href="<?=$eti['tr']?>.videos" title="<?=$eti['baslik']?>"><font color="#<? echo randchar(8,"hexcolor"); ?>"><?=$eti['baslik']?></font></a>

							



<? } ?>

				</ul>

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				<h2>Hot Videolar</h2>

<table><tr><td>En Çok izlenenler

				<ul>

				<? $rand=mysql_query("select * from madvideo order by izlenme limit 10"); 

   while ($ran=mysql_fetch_array($rand))  { ?>





							<li><a href="videos_<?=$ran['seo']?>.html" title="<?=$ran['baslik']?>"><img src="<?=$ran['resim']?>" alt="<?=$ran['baslik']?>" border="0" width="90" height="70" /></a></li>

							



<? } ?>



</ul></td><td><ul> Rastgele Videolar

<? $rand=mysql_query("select * from madvideo order by rand() limit 10"); 

   while ($ran=mysql_fetch_array($rand))  { ?>





							<li><a href="videos_<?=$ran['seo']?>.html" title="<?=$ran['baslik']?>"><img src="<?=$ran['resim']?>" border="0" alt="<?=$ran['baslik']?>" width="90" height="70" /></a></li>

							



<? } ?>











</ul></td></tr></table>

				

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tablo boşsa bazı mysql versiyonlarında uyarı veriyor engellemek için

$varmi = mysql_num_rows($sorgu);


if ($varmi > 0) {

....................


}