在SQL Server中查找重复的行

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

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

2013-11-07 08:01:59

其他回答

select orgname, count(*) as dupes, id 
from organizations
where orgname in (
    select orgname
    from organizations
    group by orgname
    having (count(*) > 1)
)
group by orgname, id

2010-01-21 20:49:49

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。

2010-01-21 20:32:49

select * from [Employees]

查找重复记录 1)使用CTE

with mycte
as
(
select Name,EmailId,ROW_NUMBER() over(partition by Name,EmailId order by id) as Duplicate from [Employees]
)
select * from mycte

2)使用GroupBy

select Name,EmailId,COUNT(name) as Duplicate from  [Employees] group by Name,EmailId

2016-11-18 12:44:38

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

2010-01-21 20:32:02

Select * from (Select orgName,id,
ROW_NUMBER() OVER(Partition By OrgName ORDER by id DESC) Rownum
From organizations )tbl Where Rownum>1

所以rowum> 1的记录将是表中的重复记录。'分区由'第一组记录，然后通过给他们序列号序列化他们。所以rownum> 1将是重复的记录，可以这样删除。

2015-03-10 05:58:02

在SQL Server中查找重复的行

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