我有一个组织的SQL Server数据库,有许多重复的行。我想运行一个选择语句来获取所有这些和被欺骗的数量,同时还返回与每个组织相关的id。

这样的陈述:

SELECT     orgName, COUNT(*) AS dupes  
FROM         organizations  
GROUP BY orgName  
HAVING      (COUNT(*) > 1)

将返回如下内容

orgName        | dupes  
ABC Corp       | 7  
Foo Federation | 5  
Widget Company | 2 

但我也想要他们的id。有什么办法可以做到吗?也许就像

orgName        | dupeCount | id  
ABC Corp       | 1         | 34  
ABC Corp       | 2         | 5  
...  
Widget Company | 1         | 10  
Widget Company | 2         | 2  

原因是还有一个单独的用户表链接到这些组织,我想把它们统一起来(因此删除dupes,用户链接到同一个组织,而不是dupe组织)。但我想手动部分,所以我不会搞砸任何事情,但我仍然需要一个语句返回所有的dupe组织的id,这样我就可以通过用户列表。


当前回答

标记为正确的解决方案不为我工作,但我发现这个答案工作得很好:获取MySql中的重复行列表

SELECT n1.* 
FROM myTable n1
INNER JOIN myTable n2 
ON n2.repeatedCol = n1.repeatedCol
WHERE n1.id <> n2.id

其他回答

select o.orgName, oc.dupeCount, o.id
from organizations o
inner join (
    SELECT orgName, COUNT(*) AS dupeCount
    FROM organizations
    GROUP BY orgName
    HAVING COUNT(*) > 1
) oc on o.orgName = oc.orgName

你可以试试这个,这对你是最好的

 WITH CTE AS
    (
    SELECT *,RN=ROW_NUMBER() OVER (PARTITION BY orgName ORDER BY orgName DESC) FROM organizations 
    )
    select * from CTE where RN>1
    go

我有一个更好的选择,把重复的记录放在一个表中

SELECT x.studid, y.stdname, y.dupecount
FROM student AS x INNER JOIN
(SELECT a.stdname, COUNT(*) AS dupecount
FROM student AS a INNER JOIN
studmisc AS b ON a.studid = b.studid
WHERE (a.studid LIKE '2018%') AND (b.studstatus = 4)
GROUP BY a.stdname
HAVING (COUNT(*) > 1)) AS y ON x.stdname = y.stdname INNER JOIN
studmisc AS z ON x.studid = z.studid
WHERE (x.studid LIKE '2018%') AND (z.studstatus = 4)
ORDER BY x.stdname

上述查询的结果显示了所有具有唯一学生id的重复名称和重复出现的次数

点击这里查看sql . sql的结果

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

Src: https://stackoverflow.com/a/59242/1465252

您可以运行以下查询并找到带有max(id)的重复项并删除这些行。

SELECT orgName, COUNT(*), Max(ID) AS dupes 
FROM organizations 
GROUP BY orgName 
HAVING (COUNT(*) > 1)

但是您必须运行这个查询几次。