在Oracle数据库表中返回给定列的重复值及其出现次数的最简单SQL语句是什么?
例如:我有一个列为JOB_NUMBER的JOBS表。如何才能知道我是否有任何重复的JOB_NUMBERs,以及它们重复了多少次?
在Oracle数据库表中返回给定列的重复值及其出现次数的最简单SQL语句是什么?
例如:我有一个列为JOB_NUMBER的JOBS表。如何才能知道我是否有任何重复的JOB_NUMBERs,以及它们重复了多少次?
当前回答
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;
其他回答
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
我知道这是一个老帖子,但这可能会帮助一些人。
如果你需要打印表的其他列,同时检查重复使用如下:
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;
如果需要,还可以在where子句中添加一些额外的过滤器。
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。
你也可以尝试这样的方法在一个表中列出所有重复的值,比如reqitem
SELECT count(poid)
FROM poitem
WHERE poid = 50
AND rownum < any (SELECT count(*) FROM poitem WHERE poid = 50)
GROUP BY poid
MINUS
SELECT count(poid)
FROM poitem
WHERE poid in (50)
GROUP BY poid
HAVING count(poid) > 1;
SELECT SocialSecurity_Number, Count(*) no_of_rows
FROM SocialSecurity
GROUP BY SocialSecurity_Number
HAVING Count(*) > 1
Order by Count(*) desc