在组[MySQL]中获取表中的顶级n行数

Question

我只想得到分类中的前3位销售产品(按事务(id) count(id)按每个类别排列的前3种产品)。我在寻找很多可能的解决方案，但没有结果。在MySQL中，它看起来有点棘手，因为不能简单地使用top()函数等等。样本数据结构如下：

+--------+------------+-----------+
|     id |category_id | product_id|
+--------+------------+-----------+
| 1      | 10         | 32        |
| 2      | 10         | 34        |
| 3      | 10         | 32        |
| 4      | 10         | 21        |
| 5      | 10         | 100       |
| 6      | 7          | 101       |
| 7      | 7          | 39        |
| 8      | 7          | 41        |
| 9      | 7          | 39        |
+--------+------------+-----------+

Answer 1

在早期版本的MySQL中，我建议使用变量：

select cp.*
from (select cp.*,
             (@rn := if(@c = category_id, @rn + 1,
                        if(@c := category_id, 1, 1)
                       )
             ) as rn
      from (select category_id, product_id, count(*) as cnt
            from mytable
            group by category_id, product_id
            order by category_id, count(*) desc
           ) cp cross join
           (select @c := -1, @rn := 0) params
     ) cp
where rn <= 3;

Answer 2

如果您运行的是MySQL 8.0，则可以为此使用窗口函数rank()：

select *
from (
    select 
        category_id,
        product_id,
        count(*) cnt,
        rank() over(partition by category_id order by count(*) desc) rn
    from mytable
    group by category_id, product_id
) t
where rn <= 3

在早期版本中，有一个选项是使用相关子查询进行筛选：

select 
    category_id,
    product_id,
    count(*) cnt
from mytable t
group by category_id, product_id
having count(*) >= (
    select count(*)
    from mytable t1
    where t1.category_id = t.category_id and t1.product_id = t.product_id
    order by count(*) desc
    limit 3, 1
)