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;

我通常使用Oracle分析函数ROW_NUMBER()。

假设您想检查构建在列(c1, c2, c3)上的唯一索引或主键的副本。 然后你将这样做，打开ROW_NUMBER()带来的行数为>1的行数的ROWID s:

Select *
From Table_With_Duplicates
Where Rowid In (Select Rowid
                  From (Select ROW_NUMBER() Over (
                                 Partition By c1, c2, c3
                                 Order By c1, c2, c3
                               ) nbLines
                          From Table_With_Duplicates) t2
                 Where nbLines > 1)

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);

我能想到的最简单的:

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

1. 解决方案

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

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。