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。

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

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

你可以这样做:

SELECT
    o.id, o.orgName, d.intCount
FROM (
     SELECT orgName, COUNT(*) as intCount
     FROM organizations
     GROUP BY orgName
     HAVING COUNT(*) > 1
) AS d
    INNER JOIN organizations o ON o.orgName = d.orgName

如果你只想返回可以删除的记录(每个记录只留下一个)，你可以使用:

SELECT
    id, orgName
FROM (
     SELECT 
         orgName, id,
         ROW_NUMBER() OVER (PARTITION BY orgName ORDER BY id) AS intRow
     FROM organizations
) AS d
WHERE intRow != 1

编辑:SQL Server 2000没有ROW_NUMBER()函数。相反，你可以使用:

SELECT
    o.id, o.orgName, d.intCount
FROM (
     SELECT orgName, COUNT(*) as intCount, MIN(id) AS minId
     FROM organizations
     GROUP BY orgName
     HAVING COUNT(*) > 1
) AS d
    INNER JOIN organizations o ON o.orgName = d.orgName
WHERE d.minId != o.id

2010-01-21 20:37:03

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

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

2013-05-20 20:27:18

有几种方法可以选择重复的行。

对于我的解决方案，首先考虑以下表格为例

CREATE TABLE #Employee
(
ID          INT,
FIRST_NAME  NVARCHAR(100),
LAST_NAME   NVARCHAR(300)
)

INSERT INTO #Employee VALUES ( 1, 'Ardalan', 'Shahgholi' );
INSERT INTO #Employee VALUES ( 2, 'name1', 'lname1' );
INSERT INTO #Employee VALUES ( 3, 'name2', 'lname2' );
INSERT INTO #Employee VALUES ( 2, 'name1', 'lname1' );
INSERT INTO #Employee VALUES ( 3, 'name2', 'lname2' );
INSERT INTO #Employee VALUES ( 4, 'name3', 'lname3' );

第一个解决方案:

SELECT DISTINCT *
FROM   #Employee;

WITH #DeleteEmployee AS (
                     SELECT ROW_NUMBER()
                            OVER(PARTITION BY ID, First_Name, Last_Name ORDER BY ID) AS
                            RNUM
                     FROM   #Employee
                 )

SELECT *
FROM   #DeleteEmployee
WHERE  RNUM > 1

SELECT DISTINCT *
FROM   #Employee

第二个解决方案:使用标识字段

SELECT DISTINCT *
FROM   #Employee;

ALTER TABLE #Employee ADD UNIQ_ID INT IDENTITY(1, 1)

SELECT *
FROM   #Employee
WHERE  UNIQ_ID < (
    SELECT MAX(UNIQ_ID)
    FROM   #Employee a2
    WHERE  #Employee.ID = a2.ID
           AND #Employee.FIRST_NAME = a2.FIRST_NAME
           AND #Employee.LAST_NAME = a2.LAST_NAME
)

ALTER TABLE #Employee DROP COLUMN UNIQ_ID

SELECT DISTINCT *
FROM   #Employee

所有解决方案的最后使用这个命令

DROP TABLE #Employee

2013-11-07 07:06:30

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

Try

SELECT orgName, id, count(*) as dupes
FROM organizations
GROUP BY orgName, id
HAVING count(*) > 1;

2014-01-22 06:29:47

我想我知道你需要什么我需要混合各种答案，我认为我得到了他想要的解决方案:

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

有了Max id，你就会得到副本的id和他要求的原件的id:

id org name , dublicate count (missing out in this case) 
id doublicate org name , doub count (missing out again because does not help in this case)

唯一可悲的是你把它写成了这种形式

id , name , dubid , name

希望它仍然有用

2014-10-01 08:13:03

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

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

2015-03-17 04:01:51

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

2015-09-15 07:01:15

如果要删除重复项:

WITH CTE AS(
   SELECT orgName,id,
       RN = ROW_NUMBER()OVER(PARTITION BY orgName ORDER BY Id)
   FROM organizations
)
DELETE FROM CTE WHERE RN > 1

2016-06-17 09:51:06

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

假设我们有一个表'Student'，有两列:

student_id int student_name varchar 记录: +------------+---------------------+ | student_id | student_name | +------------+---------------------+ | 101 | usman | | 101 | usman | | 101 | usman | | 102 | usmanyaqoob | | 103 | muhammadusmanyaqoob | | 103 | muhammadusmanyaqoob | +------------+---------------------+

现在我们希望看到重复的记录使用这个查询:

select student_name,student_id ,count(*) c from student group by student_id,student_name having c>1;

+---------------------+------------+---+
| student_name        | student_id | c |
+---------------------+------------+---+
| usman               |        101 | 3 |
| muhammadusmanyaqoob |        103 | 2 |
+---------------------+------------+---+

2018-02-09 07:28:48

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

2018-06-25 09:22:28

 /*To get duplicate data in table */

 SELECT COUNT(EmpCode),EmpCode FROM tbl_Employees WHERE Status=1 
  GROUP BY EmpCode HAVING COUNT(EmpCode) > 1

2019-04-26 14:01:56

我使用两个方法来查找重复的行。第一种方法是最著名的用分组的方法。第二种方法是使用CTE -公共表表达式。

正如@RedFilter提到的那样，这种方式也是正确的。很多时候我发现CTE方法对我也很有用。

WITH TempOrg (orgName,RepeatCount)
AS
(
SELECT orgName,ROW_NUMBER() OVER(PARTITION by orgName ORDER BY orgName) 
AS RepeatCount
FROM dbo.organizations
)
select t.*,e.id from organizations   e
inner join TempOrg t on t.orgName= e.orgName
where t.RepeatCount>1

在上面的例子中，我们通过使用ROW_NUMBER和PARTITION by找到重复出现的情况来收集结果。然后应用where子句只选择重复次数大于1的行。将所有结果收集到CTE表中，并与Organizations表进行联接。

来源:CodoBee

2020-05-05 09:40:38

在SQL Server中查找重复的行

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