在SQL Server中查找重复的行
我有一个组织的SQL Server数据库，并有许多重复的行。 我想运行一个select语句来获取所有这些和欺骗的数量，但也返回与每个组织相关联的ID。
声明如下：
SELECT orgName, COUNT(*) AS dupes FROM organizations GROUP BY orgName HAVING (COUNT(*) > 1) 会返回类似的东西
orgName | dupes ABC Corp | 7 Foo Federation | 5 Widget Company | 2
但是我也想抓住他们的ID。 有没有办法做到这一点？ 也许就像一个
orgName | dupeCount | id ABC Corp | 1 | 34 ABC Corp | 2 | 5 ... Widget Company | 1 | 10 Widget Company | 2 | 2
原因是也有单独的用户链接到这些组织，我想统一他们（因此删除欺骗，使用户链接到同一组织，而不是愚弄组织）。 但我想手动部分，所以我没有任何东西搞砸了，但我仍然需要一个声明返回所有愚蠢组织的ID，所以我可以通过用户列表。
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
您可以运行以下查询并使用max(id)查找重复项并删除这些行。
SELECT orgName, COUNT(*), Max(ID) AS dupes FROM organizations GROUP BY orgName HAVING (COUNT(*) > 1)
但是你将不得不运行这个查询几次。
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
如果您只想返回可以删除的logging（每个logging都有一个），则可以使用：
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，您可以使用：
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
标记为正确的解决scheme不适用于我，但我发现这个工作很好的答案： 在MySql中获取重复行的列表
SELECT n1.* FROM myTable n1 INNER JOIN myTable n2 ON n2.repeatedCol = n1.repeatedCol WHERE n1.id <> n2.id
你可以试试这个，对你最好
WITH CTE AS ( SELECT *,RN=ROW_NUMBER() OVER (PARTITION BY orgName ORDER BY orgName DESC) FROM organizations ) select * from CTE where RN>1 go
Select * from (Select orgName,id, ROW_NUMBER() OVER(Partition By OrgName ORDER by id DESC) Rownum From organizations )tbl Where Rownum>1
因此，rowum> 1的logging将成为表中的重复logging。 按照第一组的分区划分，然后通过给予序列号序列化它们。 所以rownum> 1将是可以删除的重复logging。
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
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
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
您有几种方法来selectduplicate rows 。
对于我的解决scheme，首先考虑这个表格
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scheme
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
Secound解决scheme：使用identity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scheme的结束都使用这个命令
DROP TABLE #Employee
如果你想删除重复项：
WITH CTE AS( SELECT orgName,id, RN = ROW_NUMBER()OVER(PARTITION BY orgName ORDER BY Id) FROM organizations ) DELETE FROM CTE WHERE RN > 1
我想我知道你需要什么我需要混合的答案，我想我得到了他想要的解决scheme：
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
有最大的id会给你dublicate的id和原来的那个是他所要求的：
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)
只有悲伤的事情，你把它拿出来在这种forms
id , name , dubid , name
希望它仍然有帮助
select * from [Employees]
查找重复logging1）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）通过使用GroupBy
select Name,EmailId,COUNT(name) as Duplicate from [Employees] group by Name,EmailId
尝试
SELECT orgName, id, count(*) as dupes FROM organizations GROUP BY orgName, id HAVING count(*) > 1;
