Question

I am currently working on speeding up a website, that is returning 300,000+ rows from a query. While I don’t think this is too much of a load on the DB server, this query is happening in a while loop depending on the number of ‘galleries’ a user has.

For example Joe has 10 galleries in his account. Each of those galleries has x number of images, which have x number of comments on those images. So the query that is currently being run…

SELECT count(*) as total 
FROM galleryimage a 
INNER JOIN imagecomments b ON a.id=b.imgId 
WHERE a.galleryId='".$row['id']."' 
AND b.note <> ''

…is looking through all the galleryimage table 334,000 rows and the imagecomments table 76,000 rows and returning the result on each gallery. The query run on a single gallery returns a result in about 578ms, but with many galleries, say 30-40 you could be looking at a page load time of 17+ secs. Any suggestions on how to deal with this issue?

I cannot change the DB architecture….

Query for gallery id

SELECT a.id, 
       a.created, 
       a.name, 
       b.clientName, 
       a.isFeatured, 
       a.views, 
       a.clientId 
FROM gallery a 
INNER JOIN client b 
ON a.clientId = b.id 
WHERE a.isTemp = 0 
AND a.clientRef = '{$clientRef}' 
AND a.finish='1'  
AND a.isArchive='0' 
ORDER BY created 
DESC

Answer 1

You can consolidate the queries and eliminate the need for looping:

SELECT 
    a.id, 
    a.created, 
    a.name, 
    b.clientName, 
    a.isFeatured, 
    a.views, 
    a.clientId,
    COALESCE(c.img_cnt, 0) AS gallery_image_count,
    COALESCE(c.comment_cnt, 0) AS gallery_comment_count
FROM 
    gallery a 
INNER JOIN 
    client b ON a.clientId = b.id 
LEFT JOIN
(
    SELECT aa.galleryId, 
           COUNT(DISTINCT aa.id) AS img_cnt, 
           COUNT(1) AS comment_cnt
    FROM galleryimage aa
    INNER JOIN imagecomments bb ON aa.id = bb.imgId
    WHERE bb.note <> ''
    GROUP BY aa.galleryId
) c ON a.id = c.galleryId
WHERE 
    a.isTemp = 0 AND
    a.clientRef = '{$clientRef}' AND
    a.finish = 1 AND
    a.isArchive = 0 
ORDER BY 
    a.created DESC