SUM（）在MySQL中不起作用：SUM（）与DISTINCT

Question

我有4个表，分别是商店，用户，评论和评级。

我希望获得相应商店的所有评论，其中包含已审核的用户详细信息以及该商店的整体评分。

我几乎完成了单一查询。但问题是，如果商店对同一用户进行多次相同评级，则将其视为单一评级。但那个评级数是正确的。

即

从该表user_id 3被评为shop_id 1为4次。因此，计数为4，total_rating为17.

我的查询是

select review.comments, users.username, count(distinct rating.id) as rating_count,
sum(distinct rating.rating) as total_rating from users 
left join review on users.id = review.user_id and review.shop_id='1' 
left join rating on users.id = rating.user_id and rating.shop_id='1' 
where review.shop_id='1' or rating.shop_id='1' 
group by users.id, review.user_id, rating.user_id, review.id

当我运行此查询时，我得到了

但是对于user_id 3，我需要total_rating 17 ..

选中此fiddle

Answer 1

您将DISTINCT置于sum( rating.rating) as total_rating,这就是为什么结果（ 12 = 17-5 ），因为它在计算总和时只包含5次。

 select review.comments, review.user_id, count(distinct rating.id) as rating_count,
    sum( rating.rating) as total_rating from users 
    left join review on users.id = review.user_id and review.shop_id='1' 
    left join rating on users.id = rating.user_id and rating.shop_id='1' 
    where review.shop_id='1' or rating.shop_id='1' 
    group by users.id, review.user_id, rating.user_id, review.id

这是SQLFiddle

示例输出： 希望这有帮助

Answer 2

尝试此操作 - 从sum(rating.rating)中删除不同内容。由于您提供了sum(distinct rating.rating)，因此忽略了用户3为商店1提供的5个。

select review.comments, users.username, count(distinct rating.id) as rating_count,
sum(rating.rating) as total_rating from users 
left join review on users.id = review.user_id and review.shop_id='1' 
left join rating on users.id = rating.user_id and rating.shop_id='1' 
where review.shop_id='1' or rating.shop_id='1' 
group by users.id, review.user_id, rating.user_id, review.id

Answer 3

首先：从表中外连接记录然后在WHERE子句中删除它们是没有意义的。使用left join review ...，您可以说：在表查看中找到匹配的记录，如果找不到，则添加空值，以便我们保留用户记录。然后使用where review.shop_id='1'说：只记录您实际在审核中找到记录的记录。所以你要解雇那些你只是痛苦不堪的记录。你的WHERE子句使你的LEFT OUTER加入仅仅INNER JOINS。

关于你的实际问题：这源于首先加入所有表，然后尝试从结果记录中获取聚合。在加入之前聚合：

select 
  rev.comments, 
  usr.username, 
  coalesce(rat.rating_count, 0) as rating_count,
  rat.total_rating 
from review rev
join users usr on users.id = review.user_id 
left join
(
  select user_id, shop_id, count(*) as rating_count, sum(rating) as total_rating
  from rating 
  group by user_id, shop_id
) rat on rat.user_id = usr.id and rat.shop_id = rev.shop_id
where rev.shop_id = 1 
group by rev.id;