Question

I’m running an MySQL query that returns results based on location. However I have noticed recently that its really slowing down my PHP app. I used CodeIgniter and the profiler shows the query taking 4.2seconds. The geoname table has 500,000 rows. I have some indexes on the key columns, how else can speed up this query?

Here is my SQL:

SELECT `products`.`product_name`
     , `geoname`.`geonameid`
     , `geoname`.`latitude`
     , `geoname`.`longitude`
     , `products`.`product_id`
     , AVG(ratings.vote) as rating
     , count(comments.comment_id) as total_comments
     ,   (6371 * acos(cos(radians(38.7666667)) 
               * cos(radians(geoname.latitude)) 
               * cos(radians(geoname.longitude) - radians(-3.3833333)) 
             +   sin(radians(38.7666667)) 
               * sin(radians(geoname.latitude)))
         ) AS distance
FROM (`foods`)
JOIN `geoname` ON `geoname`.`geonameid` = `products`.`geoname_id`
LEFT JOIN `ratings` 
  ON `ratings`.`var_id` = `products`.`product_id`
LEFT JOIN `comments` 
  ON `comments`.`var_id` = `products `.`product_id`
WHERE `products`.`product_id` != 82
GROUP BY `products`.`product_id`
HAVING `distance` < 99
ORDER BY `distance`
LIMIT 10

Answer 1

Let’s start with the query itself
the cos(radians(geoname.latitude)) and other functions seem like an invariant, so we can do a little preprocessing and store the calculated values in the table. (calculating trig functions mostly involve using a series expansio which is costly).

6371 * acos(cos(radians(38.7666667)) – this is equal to radians(38.76667) * 6371 so why us it? it costs.

Secondly if You don’t care THAT much about precision You can precalc the radians itself for let’s say 10000 points from 0 to pi/2 – that should give a nice approximation, up to 4 decimal numbers eg less than a km

(6371 * acos(cos(radians(38.7666667))
 * cos(radians(geoname.latitude))
 * cos(radians(geoname.longitude) - radians(-3.3833333))
+ sin(radians(38.7666667))
* sin(radians(geoname.latitude))))

also remember that sin(a) when a > pi/2 and a < pi equals to sin(pi – a)
when a> pi and a < 3/2 pi equals to -sin(a-pi) and when a > 3/2 pi and a < 2pi it’s equal to -sin (2pi – a). similiar functions can be made for cos function.

Try this and see if it helps.
luke