Question

我正在尝试返回一个Customers表中的所有实例，其中statustype =＆＃39; dc＆＃39;那么对于那些结果，FC的计数是> 1，地址1上的计数为1。

IE：

FC    Address1
111    abc
111    cde
432    qqq
432    qqq

我需要111 FC结果，因为他们的地址1不同。但我不需要返回432 FC结果，因为该FC的地址超过1个

 SELECT *
  FROM Customers
  where FC IN( select FC from Customers where StatusType= 'dc'
  group by FC having COUNT(FC) > 1 and COUNT(Address1) < 2

)
  order by FC, Address1

我也试过= 1而不是＆lt; 2

Answer 1

如果您需要有关具有多个唯一地址的FC的详细信息，那么此查询将为您提供：

select c.* from customers c
join (
    select FC
    from customers
    where statustype = 'dc'
    group by fc having count(distinct Address1) > 1
) a on c.FC = a.FC

Answer 2

您还需要按地址分组。

SELECT *
FROM Customers
where FC IN( select FC from Customers where StatusType= 'dc'
group by FC, Address1 having COUNT(FC) > 1 and COUNT(Address1) < 2)
order by FC, Address1

Answer 3

尝试使用Distinct COUNT

SELECT *
FROM   Customers
WHERE  FC IN(SELECT FC
             FROM   Customers
             WHERE  StatusType = 'dc'
             GROUP  BY FC
             HAVING Count(DISTINCT Address1) > 1)
ORDER  BY FC,
          Address1

选择1个字段的计数大于值的位置，另一个的计数小于值

3 个答案: