如何选择列值不明显的每一行

Question

我需要运行一个select语句,返回列的值不相同的所有行(例如EmailAddress).

例如,如果表格如下所示:

CustomerName     EmailAddress
Aaron            aaron@gmail.com
Christy          aaron@gmail.com
Jason            jason@gmail.com
Eric             eric@gmail.com
John             aaron@gmail.com

我需要查询返回:

Aaron            aaron@gmail.com
Christy          aaron@gmail.com
John             aaron@gmail.com

我阅读了很多帖子并尝试了不同的查询无济于事.我认为应该工作的查询如下.有人可以提出替代方案或告诉我查询可能有什么问题吗？

select EmailAddress, CustomerName from Customers
group by EmailAddress, CustomerName
having COUNT(distinct(EmailAddress)) > 1

Answer 1

这比以下EXISTS方式快得多:

SELECT [EmailAddress], [CustomerName] FROM [Customers] WHERE [EmailAddress] IN
  (SELECT [EmailAddress] FROM [Customers] GROUP BY [EmailAddress] HAVING COUNT(*) > 1)

Answer 2

您的查询不正确的是您通过电子邮件和名称进行分组,这组合了每组唯一的电子邮件和名称组合在一起,因此

aaron and aaron@gmail.com
christy and aaron@gmail.com
john and aaron@gmail.com

被视为3个不同的组,而不是所有属于1个单独的组.

请使用下面给出的查询:

select emailaddress,customername from customers where emailaddress in
(select emailaddress from customers group by emailaddress having count(*) > 1)

Answer 3

怎么样

SELECT EmailAddress, CustomerName FROM Customers a
WHERE Exists ( SELECT emailAddress FROM customers c WHERE a.customerName != c.customerName AND a.EmailAddress = c.EmailAddress)

Answer 4

select CustomerName,count(1) from Customers group by CustomerName having count(1) > 1

Answer 5

只是为了好玩，这是另一种方式：

;with counts as (
    select CustomerName, EmailAddress,
      count(*) over (partition by EmailAddress) as num
    from Customers
)
select CustomerName, EmailAddress
from counts
where num > 1