ORDER BY variable AFTER statement - php

Question

Here is my code:
<?php
$data = mysql_query("SELECT * FROM board") or die(mysql_error());
while($info = mysql_fetch_assoc( $data ))
{
if(!empty($info['user'])){
Print "".$info['user'].""; }
else {
}
myOtherQuery($info['id']);
}
function myOtherQuery($id) {
$result3 = mysql_query("SELECT COUNT(source_user_id) FROM likes
INNER JOIN pins ON pins.id = likes.pin_id
WHERE pins.board_id='$id'");
$c = mysql_result($result3, 0); // Cumulative tally of likes for board
{
Print "$c";
}
}
?>
The first part gets a users name and board details (board as in a photo album).
the second part joins that data with another sql table that counts the number of likes that board has.
Both are displayed as a name and a score represented by a number.
By default they are ordered by the date of creation. I'd like to be able to order them by the score. However, since the score is determined in the second part of the code, I don't know how to achieve it. Is it possible?

Answer 1

The solution is of course to query both at once in the first place, via a LEFT JOIN against a subquery returning the count per board_id:
SELECT
board.*,
/* Your PHP code will retrieve the likes count via this alias `numlikes` as in $info['numlikes'] */
numlikes
FROM
board
LEFT JOIN (
/* Subquery returns count per board_id */
SELECT pins.board_id, COUNT(source_user_id) AS numlikes
FROM
likes
INNER JOIN pins ON pins.id = likes.pin_id
GROUP BY pins.board_id
) likes ON board.id = likes.board_id
ORDER BY numlikes
It is nearly always significantly more efficient to perform a single query rather than n queries in a loop. You should strive to do so whenever possible.

Answer 2

You can do it in one query
SELECT board.*, count(likes.source_user_id) as score
FROM board
INNER JOIN pins
ON pins.board_id = board.id
INNER JOIN likes
ON pins.id = likes.pin_id
ORDER BY score