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
SELECT * FROM users u where rowid = (select max(rowid) from users u1 where
u.email=u1.email);
其他回答
SELECT name, email
FROM users
WHERE email in
(SELECT email FROM users
GROUP BY email
HAVING COUNT(*)>1)
尝试此代码
WITH CTE AS
( SELECT Id, Name, Age, Comments, RN = ROW_NUMBER()OVER(PARTITION BY Name,Age ORDER BY ccn)
FROM ccnmaster )
select * from CTE
如果您希望查看表中是否有重复的行，我使用以下查询：
create table my_table(id int, name varchar(100), email varchar(100));
insert into my_table values (1, 'shekh', 'shekh@rms.com');
insert into my_table values (1, 'shekh', 'shekh@rms.com');
insert into my_table values (2, 'Aman', 'aman@rms.com');
insert into my_table values (3, 'Tom', 'tom@rms.com');
insert into my_table values (4, 'Raj', 'raj@rms.com');
Select COUNT(1) As Total_Rows from my_table
Select Count(1) As Distinct_Rows from ( Select Distinct * from my_table) abc
我想这对你有帮助
SELECT name, email, COUNT(* )
FROM users
GROUP BY name, 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
