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

试试看:

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

其他回答

 select emp.ename, emp.empno, dept.loc 
          from emp
 inner join dept 
          on dept.deptno=emp.deptno
 inner join
    (select ename, count(*) from
    emp
    group by ename, deptno
    having count(*) > 1)
 t on emp.ename=t.ename order by emp.ename
/

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);
select name, email
, case 
when ROW_NUMBER () over (partition by name, email order by name) > 1 then 'Yes'
else 'No'
end "duplicated ?"
from users

确切的代码可能会有所不同,这取决于您是希望同时查找重复的行,还是仅查找具有相同电子邮件和名称的不同ID。如果id是主键或具有唯一约束,则不存在这种区别,但问题没有对此进行说明。在前一种情况下,您可以使用其他几个答案中给出的代码:

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

在后一种情况下,您可以使用:

SELECT name, email, COUNT(DISTINCT id)
FROM users
GROUP BY name, email
HAVING COUNT(DISTINCT id) > 1
ORDER BY COUNT(DISTINCT id) DESC

SELECT id,COUNT(id)FROM table1 GROUP BY id HAVING COUNT;

我认为这可以正确地搜索特定列中的重复值。