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;

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

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;

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

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;