SELECT   SocialSecurity_Number, Count(*) no_of_rows
FROM     SocialSecurity 
GROUP BY SocialSecurity_Number
HAVING   Count(*) > 1
Order by Count(*) desc 

下面是一个SQL请求:

select column_name, count(1)
from table
group by column_name
having count (column_name) > 1;

我能想到的最简单的:

select job_number, count(*)
from jobs
group by job_number
having count(*) > 1;

按COUNT聚合列，然后使用HAVING子句查找出现多次的值。

SELECT column_name, COUNT(column_name)
FROM table_name
GROUP BY column_name
HAVING COUNT(column_name) > 1;

我知道这是一个老帖子，但这可能会帮助一些人。

如果你需要打印表的其他列，同时检查重复使用如下:

select * from table where column_name in
(select ing.column_name from table ing group by ing.column_name having count(*) > 1)
order by column_name desc;

如果需要，还可以在where子句中添加一些额外的过滤器。

1. 解决方案

select * from emp
    where rowid not in
    (select max(rowid) from emp group by empno);