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。

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。

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)

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;

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