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

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。

下面是一个SQL请求:

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

做

select count(j1.job_number), j1.job_number, j1.id, j2.id
from   jobs j1 join jobs j2 on (j1.job_numer = j2.job_number)
where  j1.id != j2.id
group by j1.job_number

将给出复制行的id。

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

1. 解决方案

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