Question

直到今天，这是一个查询，我需要知道我的数据库中的照片：

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

但是，现在我想知道为这张照片投了多少票。

这是我的'投票'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 ;

这是我尝试修改我的SQL查询以获取投票计数：

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

上述尝试导致此错误：

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

Answer 1

错误是由语法问题引起的。这是星号。此查询可以根据需要正确执行：

SELECT 
    users.facebook_id, 
    users.first_name, 
    users.last_name, 
    photos.filename, 
    photos.description, 
    photos.finalist, 
    bookmarks.photo_id AS bookmark,
    count(votes.photo_id) 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

使用count函数时，什么会导致此查询提供此错误？

1 个答案: