I have tried many times but every time its getting sql syntax error so i splited it into two queries but i want it to be one.Please help me.Follwing is the code:
$pictures = $db->get_results(sprintf("SELECT imageID, image_name, image_date, username, userID
FROM images LEFT JOIN users ON images.image_user = users.userID
WHERE show_in_gallery = 'Y' AND image_user = '%d'
ORDER BY imageID DESC LIMIT 0, 24", $data['userID']));
$stats = $db->get_results(sprintf("SELECT STAT.imageID, STAT.image_views,STAT.unique_views,STAT.earnings
FROM image_stats as STAT LEFT JOIN images as IMG ON STAT.imageID = IMG.imageID
WHERE image_user = '%d'
ORDER BY imageID DESC LIMIT 0, 24", $data['userID']));
And here is the error:
Warning: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'is ON is.imageID = i.imageID WHERE show_in_gallery= 'Y' AND image_user = '8' ORD' at line 1
Probably what you want? Not entirely sure why your attempt failed.
SELECT IMG.imageID, IMG.image_name, IMG.image_date,
USR.username, USR.userID,
STAT.image_views, STAT.unique_views, STAT.earnings
FROM image_stats as STAT LEFT JOIN images as IMG
ON STAT.imageID = IMG.imageID
LEFT JOIN users as USR
ON IMG.image_user = USR.userID
WHERE image_user = '%d' AND USR.show_in_gallery = 'Y'
ORDER BY imageID DESC LIMIT 0, 24
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.