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

删除名称重复的记录

;WITH CTE AS    
(

    SELECT ROW_NUMBER() OVER (PARTITION BY name ORDER BY name) AS T FROM     @YourTable    
)

DELETE FROM CTE WHERE T > 1

其他回答

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

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

派对晚了一点,但我找到了一个非常酷的解决方法来查找所有重复的ID:

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

另一种简单的方法是使用解析函数:

SELECT * from 

(SELECT name, email,

COUNT(name) OVER (PARTITION BY name, email) cnt 

FROM users)

WHERE cnt >1;

试试看:

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