如何在Oracle中的表中找到重复值？

Question

什么是最简单的SQL语句，它将返回给定列的重复值及其在Oracle数据库表中的出现次数？

例如：我有一个qazxsw poi表，列qazxsw poi。我怎样才能知道我是否有任何重复的JOBSs，以及它们重复了多少次？

Answer 1

JOB_NUMBER

Answer 2

这是执行此操作的SQL请求：

>1

Answer 3

我知道这是一个旧线程，但这可能对某人有所帮助。

如果您需要打印表格的其他列，同时检查下面的重复使用：

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

如果需要，还可以在where子句中添加一些额外的过滤器。

Answer 4

解决方案

select column_name, count(1)
from table
group by column_name
having count (column_name) > 1;

Answer 5

你也可以尝试这样的方法来列出表格中的所有重复值，例如reqitem

select * from table where column_name in
(select ing.column_name from table ing group by ing.column_name having count(*) > 1)
order by column_name desc;

Answer 6

其他方式：

JOB_NUMBER

当

SELECT column_name, COUNT(column_name)
FROM table_name
GROUP BY column_name
HAVING COUNT(column_name) > 1;

上有索引时工作正常（足够快）。并且它是删除或更新重复行的更好方法。

Answer 7

最简单的我能想到：

SELECT *
FROM TABLE A
WHERE EXISTS (
  SELECT 1 FROM TABLE
  WHERE COLUMN_NAME = A.COLUMN_NAME
  AND ROWID < A.ROWID
)

Answer 8

如果您不需要知道重复的实际数量，则不需要在返回的列中包含计数。例如

column_name

Answer 9

怎么样：

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

要回答上面的例子，它看起来像：

SELECT column_name
FROM table
GROUP BY column_name
HAVING COUNT(*) > 1

Answer 10

如果多列标识唯一行（例如关系表），您可以使用以下内容

使用行ID，例如emp_dept（empid，deptid，startdate，enddate）假设empid和deptid是唯一的并且在那种情况下识别行

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

如果这样的表有主键，那么使用主键而不是rowid，例如id是pk然后

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

Answer 11

干

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);

会给你重复的行ID。

Answer 12

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);

Answer 13

我通常使用

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

函数

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

。

假设您要检查有关在列（Oracle Analytic，ROW_NUMBER()，c1）上构建的唯一索引或主键的重复项。然后你会这样，提出c2的行，其中c3带来的行数是ROWID：

ROW_NUMBER()