Question

我有一个“bloomp”（就像一个帖子），当我从数据库中选择它时，我会记录评论的数量，喜欢以及当前用户是否喜欢。

问题在于喜欢和评论的数量。例如：

如果bloomp有3个喜欢和1个评论，则选择返回它有3个喜欢和3个评论，反之亦然。

这是我的疑问：

SELECT `bloomps`.`id` AS id,
       `bloomps`.`text` AS text,
       `bloomps`.`date` AS date,
       `images`.`id` AS image_id,
       `videos`.`id` AS video_id,
       `videos`.`provider` AS video_provider,
       `users`.`name` AS author_name,
       `users`.`picture` AS author_picture,
       SUM(CASE WHEN likes.bloomp_id IS NOT NULL THEN 1 ELSE 0 END) AS likes_number,
       SUM(CASE WHEN likes.bloomp_id IS NOT NULL AND likes.author_id = 1 THEN 1 ELSE 0 END) AS liked,
       SUM(CASE WHEN comments.bloomp_id IS NOT NULL THEN 1 ELSE 0 END) AS comments_number
FROM (`bloomps`)
INNER JOIN `users` ON `users`.`id` = `bloomps`.`author_id`
LEFT JOIN `likes` ON `likes`.`bloomp_id` = `bloomps`.`id`
LEFT JOIN `comments` ON `comments`.`bloomp_id` = `bloomps`.`id`
LEFT JOIN `images` ON `images`.`bloomp_id` = `bloomps`.`id`
LEFT JOIN `videos` ON `videos`.`bloomp_id` = `bloomps`.`id`
GROUP BY `bloomps`.`id`
ORDER BY `bloomps`.`id` ASC
LIMIT 10

enter image description here

此查询：

SELECT `bloomps`.`id` AS id,
       `bloomps`.`text` AS text,
       `bloomps`.`date` AS date,
       `users`.`name` AS author_name,
       `users`.`picture` AS author_picture,
       SUM(CASE WHEN likes.bloomp_id IS NOT NULL THEN 1 ELSE 0 END) AS likes_number,
       SUM(CASE WHEN likes.bloomp_id IS NOT NULL AND likes.author_id = 1 THEN 1 ELSE 0 END) AS liked,
       SUM(CASE WHEN comments.bloomp_id IS NOT NULL THEN 1 ELSE 0 END) AS comments_number
FROM (`bloomps`)
INNER JOIN `users` ON `users`.`id` = `bloomps`.`author_id`
LEFT JOIN `likes` ON `likes`.`bloomp_id` = `bloomps`.`id`
LEFT JOIN `comments` ON `comments`.`bloomp_id` = `bloomps`.`id`
GROUP BY `bloomps`.`id`
ORDER BY `bloomps`.`id` ASC
LIMIT 10

回归：

Answer 1

其他联接正在弄乱你的数量。例如，假设一个 bloomp 有两个视频和两个评论;如目前所写，您的查询将返回4行，使其看起来有4条评论。要解决此问题，请删除所有其他内容（请参阅下文），或将其添加到Group By子句中。

这应该为您提供基本信息 - ID，以及评论和喜欢：

SELECT `bloomps`.`id` AS id,
   SUM(CASE WHEN likes.bloomp_id IS NOT NULL THEN 1 ELSE 0 END) AS likes_number,
   SUM(CASE WHEN comments.bloomp_id IS NOT NULL THEN 1 ELSE 0 END) AS comments_number
FROM (`bloomps`)
LEFT JOIN `likes` ON `likes`.`bloomp_id` = `bloomps`.`id`
LEFT JOIN `comments` ON `comments`.`bloomp_id` = `bloomps`.`id`
GROUP BY `bloomps`.`id`
ORDER BY `bloomps`.`id` ASC
LIMIT 10

Answer 2

您已将多个表连接到主（bloomps）表，并且所有这些表都与它具有1对多的关系。结果是小型笛卡尔连接和分组时的错误计数。

您需要在子查询中单独对每个组进行分组，然后将它们连接到基表 - 并且不在那里进行分组。试试这个：

SELECT 
    b.id
  , b.*                                -- whatever column from `bloomps`
  , COALESCE(l.cnt,0)   AS likes_number,
  , COALESCE(l.liked,0) AS liked,
  , COALESCE(c.cnt,0)   AS comments_number
FROM 
      bloomps AS b
   LEFT JOIN 
      ( SELECT bloomp_id
             , COUNT(*) AS cnt 
             , COUNT(CASE WHEN author_id = 1 THEN 1 END) AS liked
        FROM likes
        GROUP BY bloomp_id
      ) AS l
         ON l.bloomp_id = b.id
   LEFT JOIN 
      ( SELECT bloomp_id
             , COUNT(*) AS cnt 
        FROM comments
        GROUP BY bloomp_id
      ) AS c
         ON c.bloomp_id = b.id
                                         -- no need for GROUP BY
ORDER BY id ASC 
  LIMIT 10 ;

如果您不想要子查询，可以使用以下修改来保留查询。测试哪一个更有效：

SELECT 
    b.id
  , b.test
  , b.date 
  , COUNT(DISTINCT l.id)  AS likes_number,
  , COUNT(DISTINCT CASE WHEN l.author=1 THEN l.id END) AS liked,
  , COUNT(DISTINCT c.id)  AS comments_number
FROM 
      bloomps AS b
   LEFT JOIN 
      likes AS l
         ON l.bloomp_id = b.id
   LEFT JOIN 
      comments AS c
         ON c.bloomp_id = b.id
GROUP BY b.id ASC                  -- MySQL non-standard syntax for
                                   -- GROUP BY b.id 
                                   -- ORDER BY b.id ASC 
  LIMIT 10 ;

Answer 3

更新：

SELECT table1.id, table1.likes_number, table2.comments_number FROM
    (SELECT b.id AS id, SUM(IF(l.bloomp_id IS NULL,0,1) 
     FROM bloomps b LEFT JOIN likes l ON b.id = l.bloomp_id GROUP BY b.id) AS table1
JOIN
    (SELECT b.id AS id, SUM(IF(c.bloomp_id IS NULL,0,1))
     FROM bloomps b LEFT JOIN comments c ON b.id = c.bloomp_id GROUP BY b.id) AS table2
ON table1.id = table2.id 
ORDER BY table1.id ASC
LIMIT 10;

MySQL多个SUM返回相同的值

3 个答案: