如何删除没有唯一行id存在的重复行?
我的座位是
col1 col2 col3 col4 col5 col6 col7
john 1 1 1 1 1 1
john 1 1 1 1 1 1
sally 2 2 2 2 2 2
sally 2 2 2 2 2 2
我想留下以下重复删除后:
john 1 1 1 1 1 1
sally 2 2 2 2 2 2
我尝试了一些查询，但我认为他们取决于有一个行id，因为我没有得到想要的结果。例如:
DELETE
FROM table
WHERE col1 IN (
SELECT id
FROM table
GROUP BY id
HAVING (COUNT(col1) > 1)
)
当前回答
DECLARE @TB TABLE(NAME VARCHAR(100));
INSERT INTO @TB VALUES ('Red'),('Red'),('Green'),('Blue'),('White'),('White')
--**Delete by Rank**
;WITH CTE AS(SELECT NAME,DENSE_RANK() OVER (PARTITION BY NAME ORDER BY NEWID()) ID FROM @TB)
DELETE FROM CTE WHERE ID>1
SELECT NAME FROM @TB;
--**Delete by Row Number**
;WITH CTE AS(SELECT NAME,ROW_NUMBER() OVER (PARTITION BY NAME ORDER BY NAME) ID FROM @TB)
DELETE FROM CTE WHERE ID>1;
SELECT NAME FROM @TB;
其他回答
DELETE p1 FROM Person p1,
Person p2
WHERE
p1.Email = p2.Email AND p1.Id > p2.Id
这可能对你的情况有帮助
DELETE t1 FROM table t1 INNER JOIN table t2 WHERE t1.id > t2.id AND t1.col1 = t2.col1
DECLARE @TB TABLE(NAME VARCHAR(100));
INSERT INTO @TB VALUES ('Red'),('Red'),('Green'),('Blue'),('White'),('White')
--**Delete by Rank**
;WITH CTE AS(SELECT NAME,DENSE_RANK() OVER (PARTITION BY NAME ORDER BY NEWID()) ID FROM @TB)
DELETE FROM CTE WHERE ID>1
SELECT NAME FROM @TB;
--**Delete by Row Number**
;WITH CTE AS(SELECT NAME,ROW_NUMBER() OVER (PARTITION BY NAME ORDER BY NAME) ID FROM @TB)
DELETE FROM CTE WHERE ID>1;
SELECT NAME FROM @TB;
不使用CTE和ROW_NUMBER()，您可以通过使用group by和MAX函数删除记录，这里是一个例子
DELETE
FROM MyDuplicateTable
WHERE ID NOT IN
(
SELECT MAX(ID)
FROM MyDuplicateTable
GROUP BY DuplicateColumn1, DuplicateColumn2, DuplicateColumn3)
在sql server中可以通过多种方式来实现 最简单的方法是: 将重复行表中的不同行插入到新的临时表中。然后从重复的行表中删除所有数据，然后从没有重复的临时表中插入所有数据，如下所示。
select distinct * into #tmp From table
delete from table
insert into table
select * from #tmp drop table #tmp
select * from table
使用公共表表达式(CTE)删除重复行
With CTE_Duplicates as
(select id,name , row_number()
over(partition by id,name order by id,name ) rownumber from table )
delete from CTE_Duplicates where rownumber!=1
