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

这也应该奏效,也许试试看。

  Select * from Users a
            where EXISTS (Select * from Users b 
                where (     a.name = b.name 
                        OR  a.email = b.email)
                     and a.ID != b.id)

如果您搜索具有某种前缀或一般更改(例如邮件中的新域)的重复项,则特别适合您。然后可以在这些列中使用replace()

其他回答

您可以使用SELECT DISTINCT关键字来消除重复项。您还可以按名称进行筛选,并在表中获取具有该名称的所有人。

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 

从表中的重复记录检查。

select * from users s 
where rowid < any 
(select rowid from users k where s.name = k.name and s.email = k.email);

or

select * from users s 
where rowid not in 
(select max(rowid) from users k where s.name = k.name and s.email = k.email);

删除表中的重复记录。

delete from users s 
where rowid < any 
(select rowid from users k where s.name = k.name and s.email = k.email);

or

delete from users s 
where rowid not in 
(select max(rowid) from users k where s.name = k.name and s.email = k.email);

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

在使用Microsoft Access的情况下,此方法有效:

CREATE TABLE users (id int, name varchar(10), email varchar(50));

INSERT INTO users VALUES (1, 'John', 'asd@asd.com');
INSERT INTO users VALUES (2, 'Sam', 'asd@asd.com');
INSERT INTO users VALUES (3, 'Tom', 'asd@asd.com');
INSERT INTO users VALUES (4, 'Bob', 'bob@asd.com');
INSERT INTO users VALUES (5, 'Tom', 'asd@asd.com');

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

DELETE *
FROM users
WHERE id IN (
    SELECT u1.id 
    FROM users u1, users u2 
    WHERE u1.name = u2.name AND u1.email = u2.email AND u1.id > u2.id
);

感谢Tancrede Chazallet的删除代码。