[英]SUM() not working in MySQL : SUM() with DISTINCT
我有4个表,称为商店,用户,评论和评级。
我想获得相应商店的所有评论,包括评论的用户详细信息以及该商店的整体评分。
我几乎完成了单个查询。 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。 但该评级数是正确的。
即
从该表中,user_id 3被评为shop_id 1为4次。 因此计数为4,total_rating为17。
我的疑问是
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
但我需要total_rating 17 for user_id 3 ..
检查这个小提琴
你把DISTINCT
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
试试这个 - 删除与sum(rating.rating)
的区别。 由于你给了sum(distinct rating.rating)
,它忽略了用户3给商店1的5。
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
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 ...
你说:在表评论中找到匹配的记录,如果你没有找到任何,那么添加空值,所以我们保留用户记录。 然后where review.shop_id='1'
你说:只记录你实际在审查中找到记录的记录。 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。
至于你的实际问题:这源于首先加入所有表,然后尝试从结果记录中获取聚合。 在加入之前汇总:
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;
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