我能想到的最简单的:

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

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

在多个列标识唯一行的情况下(例如关系表)，你可以使用以下

使用行id 例如emp_dept(empid, deptid，开始日期，结束日期) 假设empid和deptid是唯一的，并在这种情况下标识行

select oed.empid, count(oed.empid) 
from emp_dept oed 
where exists ( select * 
               from  emp_dept ied 
                where oed.rowid <> ied.rowid and 
                       ied.empid = oed.empid and 
                      ied.deptid = oed.deptid )  
        group by oed.empid having count(oed.empid) > 1 order by count(oed.empid);

如果这样的表有主键，那么使用主键而不是rowid，例如id是pk那么

select oed.empid, count(oed.empid) 
from emp_dept oed 
where exists ( select * 
               from  emp_dept ied 
                where oed.id <> ied.id and 
                       ied.empid = oed.empid and 
                      ied.deptid = oed.deptid )  
        group by oed.empid having count(oed.empid) > 1 order by count(oed.empid);

1. 解决方案

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

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;