Question

我必须每周从Customer（客户名称，ID）和order（订单ID，订单日期，订单日期）中找到3个最高消费客户。如果我今天运行查询，它应显示订单日期存在的所有周的前3位。

我正在考虑在约会（每周）上做Partition by，但我找不到任何方法来做到这一点？有人做过每周一次的结果分区吗？

我知道这不对，但这就是我所拥有的：

Select Top 3 customer_name, id OVER (partition by [week])
(
    Select c.customer_name, c.id, o.order_amt,
    from customer c
    Join Order o
        on c.id=o.id
    group by c.id 
)

Answer 1

查询应该是这样的

O(n)

Answer 2

根据您的表格结构，orders.order_id为customer.id 使用此声明

select
    *
from
(
    select
        details.*
        ,dense_rank() over (partition by week_num order by order_amt desc) as rank_num
    from
    (
        select
            c.id as customer_id
            ,c.name
            ,sum(o.order_amt) as order_amt
            ,datepart(WEEK,o.order_date) as week_num    
        from customer c
        join orders o on c.id=o.order_id
        group by c.id,c.name,datepart(WEEK,o.order_date)
    )details

)dets 
where dets.rank_num<=3

更新：已更改语句仅使用2个表

Answer 3

这是一个想法，

;WITH CTE
AS (
    SELECT c.customer_name
        ,c.id
        ,o.order_amt
        ,datepart(wk, datecol) AS Weekcol
    )
    ,CTE1
AS (
    SELECT c.customer_name
        ,c.id
        ,o.order_amt
        ,ROW_NUMBER() OVER (
            PARTITION BY Weekcol ORDER BY order_amt DESC
            ) AS rowNUm
    FROM CTE
    )
SELECT *
FROM CTE1
WHERE rowNUm <= 3

如何在SQL中每周对结果进行分区？

3 个答案: