Mysql 按提交总数选择用户

我知道这很简单，但它让我发疯

我有一个用户表、评论表和图片表

我想要一个名单的前10名用户的基础上提交的意见和他们提交的照片总数

就这样

真丢人

更新：根据Ed的回答

以下是我的设置：

用户表用户id、用户名 图像表img_id，提交者为_id=users.user_id 注释表id，由用户id=users.user\u id提交 最后一个问题是：

    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;

---

psuedocode当然可以，但您需要以下内容：

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

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;

---

也许是这样的：

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

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：

统计每个表中用户提交的内容 合并这些，这样每个用户都将从子查询中获得0到2个计数。 再分组一次，将两个计数相加，然后按顺序排列，使最高数量位于顶部。 希望这有帮助。

调整Ed的答案：

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

---

我相信你想在这里进行联合，只要联合就可以忽略看起来相同id和提交计数的记录。

总提交计数将是错误的，因为评论和图片之间存在笛卡尔积。不过，如果只需要对顶级用户进行排序，那也没关系。啊，你说得对。Ed的帖子实际上是更准确的方法。1248-每个派生表都必须有自己的别名。。。我是不是在什么地方漏了一个as z？是的，在右括号后面和groupby之前加上一个as x。这是这里的正确答案。