mysql - 选择给定日期每组的前 10 行

Question

我有一个这样的表:

id int primary key
user_id int, 
activity_date date, 
activity_type int, 
activity_value float

uniq 键(user_id、activity_date、activity_type)

我想获取给定日期 (activity_date)(例如今天或昨天)每个 activity_type 中按 activity_value 排序的前 10 行.

我尝试了自连接变体，但没有得到正确的结果。这是我尝试过的。

select 
  a.user_id, a.activity_date, a.activity_type, a.activity_value
from 
  my_table a
left join 
  my_table b on a.activity_type = b.activity_type and a.id < b.id
where 
  a.activity_date = curdate()
group by 
  a.id
having 
  count(*) < 10
order by 
  a.activity_value desc;

SQL fiddle :http://sqlfiddle.com/#!9/d30332/3

有人可以帮我纠正这个查询吗？

Answer 1

好吧，我想你已经差不多了，我想你只是错过了 b.activity_date = curdate() 条件。另外我认为LEFT JOIN是不必要的。

select a.*
from my_table a
join my_table b on a.activity_type = b.activity_type and 
                   a.activity_value <= b.activity_value
where a.activity_date = curdate() and b.activity_date = curdate()
group by a.id
having count(*) <= 10

demo