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。

下面是一个SQL请求:

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

1. 解决方案

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

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

如果您不需要知道重复的实际数量，则甚至不需要在返回列中显示计数。如。

SELECT column_name
FROM table
GROUP BY column_name
HAVING COUNT(*) > 1

另一种方法:

SELECT *
FROM TABLE A
WHERE EXISTS (
  SELECT 1 FROM TABLE
  WHERE COLUMN_NAME = A.COLUMN_NAME
  AND ROWID < A.ROWID
)

当column_name上有索引时，工作正常(足够快)。它是删除或更新重复行的更好方法。