Distinct and Ordered Results from Average PHP - php

Question

I'm trying to get the averages from the rating column of my table, then display a row in a HTML table for each distinct div_id and its average rating, in descending order. I know this should probably be easy, but I'm having a hard time figuring it out. Any help would be greatly appreciated.
<?php
mysql_connect($db_server, $db_username, $db_password) or die(mysql_error());
mysql_select_db($db_database) or die(mysql_error());
$result = mysql_query("SELECT * FROM ratings") or die(mysql_error());
while($row = mysql_fetch_object( $result )){
$ad = $row->div_id;
$result2 = mysql_query("SELECT div_id, avg(rating) AS avg, COUNT(*) FROM ratings WHERE div_id = '" . $ad . "' ORDER BY div_id DESC ") or die(mysql_error());
while($row2 = mysql_fetch_array( $result2 )){
$adid = $row2[0];
$count = $row2[2];
$avg = round($row2[1],2);
echo "<tr><td>";
echo $adid;
echo "</td><td>";
echo $avg;
echo "</td></tr>";
}
}
?>

Answer 1

Instead do tow queries, you can use the GROUP BY term in only one query.
SELECT div_id, avg(rating) AS avg, COUNT(*) FROM ratings GROUP BY div_id ORDER BY div_id DESC
php code
<?php
mysql_connect($db_server, $db_username, $db_password) or die(mysql_error());
mysql_select_db($db_database) or die(mysql_error());
$result = mysql_query("SELECT div_id, avg(rating) AS avg, COUNT(*) FROM ratings GROUP BY div_id ORDER BY avg DESC") or die(mysql_error());
while($row = mysql_fetch_array( $result )){
$adid = $row[0];
$count = $row[2];
$avg = round($row[1],2);
echo "<tr><td>" . $adid . "</td><td>" . $avg . "</td></tr>";
}
?>

Answer 2

Perhaps you should try doing this all in one query:
SELECT div_id, avg(rating) AS avgrating, COUNT(*) as cnt
FROM ratings
GROUP BY div_id
ORDER BY avgrating desc;