0
直到今天,這是得到的一切查詢,我需要知道從我的數據庫中的照片:當使用count函數時,會導致此查詢提供此錯誤的原因是什麼?
SELECT
users.facebook_id,
users.first_name,
users.last_name,
photos.filename,
photos.description,
photos.finalist,
bookmarks.photo_id AS bookmark
FROM `photos`, `users`
LEFT JOIN bookmarks ON bookmarks.photo_id = 123 AND bookmarks.facebook_id = 123456789
WHERE
users.facebook_id = photos.author AND
photos.id = 123
LIMIT 1
不過,現在我想也找出有多少票投給了這個照片。
這裏是我的「票」表:
CREATE TABLE IF NOT EXISTS `votes` (
`photo_id` int(11) NOT NULL,
`facebook_id` bigint(20) NOT NULL COMMENT 'The user''s Facebook ID.',
`date` varchar(10) NOT NULL COMMENT 'Date formatted as YYYY-MM-DD.',
UNIQUE KEY `one_vote_per_day` (`photo_id`,`facebook_id`,`date`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=1 ;
這裏是我試圖修改我的SQL查詢來獲取投票數:
SELECT
users.facebook_id,
users.first_name,
users.last_name,
photos.filename,
photos.description,
photos.finalist,
bookmarks.photo_id AS bookmark,
count(votes.*) AS vote_count
FROM `photos`, `users`
LEFT JOIN votes ON votes.photo_id = 123
LEFT JOIN bookmarks ON bookmarks.photo_id = 123 AND bookmarks.facebook_id = 123456789
WHERE
users.facebook_id = photos.author AND
photos.id = 123
LIMIT 1
上述嘗試導致了這個錯誤:
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 '*) AS vote_count FROM `photos`, `users` LEFT JOIN votes ON votes.photo_id = 1' at line 9
您應該接受一些回答您的其他問題(使用綠色複選標記)工作 –