Question

我的数据是这样的：

Client Number | Order Date | Order Amount | Sequence (created with Row_Number())

我用Row_Number（）创建了一个序列，所以我可以看到客户端有多少订单。

如果我使用WHERE序列＆gt; 3，我在3之前丢失了订单。我无法使用HAVING，因为我需要查看每个订单。如何选择超过3个订单的客户编号？

我想看看：

Client Number | Order Date | Order Amount | Sequence
1111             Jan 01      100             1
1111             Jan 02      100             2
1111             Jan 03      100             3
1112             Jan 01      100             1
1112             ...         ...             ...
1112             Jan 20      100             20

所以只有序列大于3的那些，同时仍然保持序列1和2的行。

Answer 1

SELECT *
  FROM data
 WHERE ClientNumber IN 
 (
   SELECT ClientNumber
     FROM data
   GROUP BY ClientNumber
   HAVING COUNT(1) >= 3
 );

Answer 2

create table #test(clientnumber int, orderdate datetime, orderamount int)

insert into #test values
(1110, '01/01/2016', 100),
(1110, '01/02/2016', 100),
(1111, '01/01/2016', 100),
(1111, '01/02/2016', 100),
(1111, '01/03/2016', 100),
(1112, '01/01/2016', 100),
(1112, '01/02/2016', 100),
(1112, '01/03/2016', 100),
(1112, '01/04/2016', 100);

with cte as(
select clientnumber, orderdate, orderamount,
count(*) over(partition by clientnumber ) as ran
from #test)
select * from cte
where ran >= 3

仅在超过X次出现时选择

2 个答案: