使用一个字段很容易找到重复项:

SELECT email, COUNT(email) 
FROM users
GROUP BY email
HAVING COUNT(email) > 1

所以如果我们有一张桌子

ID   NAME   EMAIL
1    John   asd@asd.com
2    Sam    asd@asd.com
3    Tom    asd@asd.com
4    Bob    bob@asd.com
5    Tom    asd@asd.com

这个查询将告诉我们John、Sam、Tom和Tom,因为他们都有相同的电子邮件。

然而,我想要的是获得相同电子邮件和名称的副本。

也就是说,我想得到“汤姆”,“汤姆”。

我需要这个的原因是:我犯了一个错误,允许插入重复的名称和电子邮件值。现在我需要删除/更改重复项,所以我需要先找到它们。


当前回答

我想这对你有帮助

SELECT name, email, COUNT(* ) 
FROM users
GROUP BY name, email
HAVING COUNT(*)>1

其他回答

表结构:

ID   NAME   EMAIL
1    John   asd@asd.com
2    Sam    asd@asd.com
3    Tom    asd@asd.com
4    Bob    bob@asd.com
5    Tom    asd@asd.com

解决方案1:

SELECT *,
       COUNT(*)
FROM users t1
INNER JOIN users t2
WHERE t1.id > t2.id
  AND t1.name = t2.name
  AND t1.email=t2.email

解决方案2:

SELECT name,
         email,
       COUNT(*)
FROM users
GROUP BY name,
         email
HAVING COUNT(*) > 1

您使用我使用的以下查询:

   select *
        FROM TABLENAME
        WHERE PrimaryCoumnID NOT IN
        (
            SELECT MAX(PrimaryCoumnID)
            FROM  TABLENAME
            GROUP BY AnyCoumnID
        );

你可能想试试这个

SELECT NAME, EMAIL, COUNT(*)
FROM USERS
GROUP BY 1,2
HAVING COUNT(*) > 1

通过使用CTE,我们也可以找到类似这样的重复值

with MyCTE
as
(
select Name,EmailId,ROW_NUMBER() over(PARTITION BY EmailId order by id) as Duplicate from [Employees]

)
select * from MyCTE where Duplicate>1

试试看:

SELECT name, email
FROM users
GROUP BY name, email
HAVING ( COUNT(*) > 1 )