Question

我需要找到所有附有重复客户编号的订单。

 SalesNo           CustNo
  112233              56
  998877              12
  998877              33
  110044              312

它应该返回以下结果：

 SalesNo      CustNo
 998877           12
 998877           33

我尝试了以下脚本，但我认为这只会让我得到重复输入的具有相同数据的订单。

select salesno, custno, COUNT(*)
from sales
GROUP BY salesno, custno
HAVING COUNT(*) > 1

Answer 1

您可以比较having子句中的值：

select salesno, COUNT(*)
from sales
group by salesno
having min(custno) <> max(custno)

Answer 2

我知道了。发生了以下情况：

select a.salesno, a.custno
from sales a
LEFT JOIN(
select salesno, custno
from sales
)b ON a.salesno = b.salesno
WHERE a.salesno = b.salesno AND a.custno <> b.custno
ORDER BY a.salesno

Answer 3

您可以使用DISTINCT关键字：

select salesno, custno, COUNT(*)
from sales
GROUP BY salesno, custno
HAVING COUNT(DISTINCT custno) > 1

Answer 4

另一种选择是将WITH TIES与sum() over()配合使用

示例

Select Top 1 with ties *
 from YourTable
 Order By sign(sum(1) over (partition by SalesNo) - 1) Desc

返回

SalesNo CustNo
998877  12
998877  33

Answer 5

您必须首先找到重复的订单号

select salesno
from sales
GROUP BY salesno
HAVING COUNT(*) > 1

然后搜索这些订单的记录

WITH duble AS (
    select salesno
    from sales
    GROUP BY salesno
    HAVING COUNT(*) > 1
)

SELECT s.* 
FROM sales s
INNER JOIN duble d ON s.salesno=d.salesno

Answer 6

使用cte（custno，cnt）为（选择11235,12所有选择11235,14所有选择11236,22全部选择11237,33）从Custno所在的Cte中选择*（从CTE组中按CUSTNO的COUNT（*）> 1选择Custno）

获取具有重复的客户编号附加到单个订单的条目

6 个答案: