Question

我知道这很容易，但它让我疯了......

我有一个用户表，评论表和图片表。

我想根据提交内容排列前十名用户（他们的评论总数和提交的照片）。

就是这样。

让我感到羞耻。

更新：根据Ed的回答。

这是我的设置：

用户表（user_id，用户名）
图像表（img_id，submittedby_id = users.user_id）
评论表（id，submittedby_id = users.user_id）

和最终查询：

    select submittedby_id, sum(total)
from 
    (select submittedby_id, count(img_id) as total from    
          images group by submittedby_id 
     union 
     select submittedby_id, count(id) as total from 
          comments group by submittedby_id
    ) as x
 group by submittedby_id 
 order by sum(total) desc limit 10;

Answer 1

也许有点像这样：

select username, sum(submissions) 
from 
    (select username, count(picture_id) from    
          pictures group by username 
     union 
     select username, count(comment_id) from 
          comments group by username
    )
 group by username 
 order by sum(submissions) desc limit 10;

从概念上概述：

计算每个表中用户的提交
联合那些，因此每个用户将从子查询中获得0到2个计数。
再组一次，将两个计数相加，然后订购，以便最高金额在最高位置。

希望这有帮助。

Answer 2

当然是伪代码，但你想要这样的东西：

select 
  u.userid
, count(commentID) + count(photoID) as totalsubmissions 
from users u
left outer 
    join comments c
    on u.userid = c.userid 
left outer 
    join pictures p 
    on u.userid = p.userid 
group by 
    u.userid 
order by 2 desc 
fetch first 10 rows only

Answer 3

调整Ed的回答：

select submittedby_id, sum(submissions) 
from 
    (select submittedby_id, count(img_id) as submissions from    
          images group by submittedby_id 
     union all
     select submittedby_id, count(id) as submissions from 
          comments group by submittedby_id
    ) as x
 group by submittedby_id 
 order by sum(submissions) desc limit 10

我相信你想在这里做一个联盟，只需联盟可以省略看起来相同的记录（相同的id和提交数）。

按总提交选择用户

3 个答案: