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

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。

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

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

如何:

SELECT <column>, count(*)
FROM <table>
GROUP BY <column> HAVING COUNT(*) > 1;

要回答上面的例子，它看起来像:

SELECT job_number, count(*)
FROM jobs
GROUP BY job_number HAVING COUNT(*) > 1;

我能想到的最简单的:

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