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
-- this query will keep only one instance of a duplicate record.
;WITH cte
AS (SELECT ROW_NUMBER() OVER (PARTITION BY col1, col2, col3-- based on what? --can be multiple columns
ORDER BY ( SELECT 0)) RN
FROM Mytable)
delete FROM cte
WHERE RN > 1
其他回答
我更喜欢CTE从sql server表中删除重复的行
强烈推荐阅读本文::http://codaffection.com/sql-server-article/delete-duplicate-rows-in-sql-server/
保持原创性
WITH CTE AS
(
SELECT *,ROW_NUMBER() OVER (PARTITION BY col1,col2,col3 ORDER BY col1,col2,col3) AS RN
FROM MyTable
)
DELETE FROM CTE WHERE RN<>1
不保留原创
WITH CTE AS
(SELECT *,R=RANK() OVER (ORDER BY col1,col2,col3)
FROM MyTable)
DELETE CTE
WHERE R IN (SELECT R FROM CTE GROUP BY R HAVING COUNT(*)>1)
删除所有重复项，但删除第一个重复项(具有最小ID)
应该同样适用于其他SQL服务器，如Postgres:
DELETE FROM table
WHERE id NOT IN (
select min(id) from table
group by col1, col2, col3, col4, col5, col6, col7
)
DELETE FROM TBL1 WHERE ID IN
(SELECT ID FROM TBL1 a WHERE ID!=
(select MAX(ID) from TBL1 where DUPVAL=a.DUPVAL
group by DUPVAL
having count(DUPVAL)>1))
另一种在不丢失信息的情况下删除重复行的方法如下:
delete from dublicated_table t1 (nolock)
join (
select t2.dublicated_field
, min(len(t2.field_kept)) as min_field_kept
from dublicated_table t2 (nolock)
group by t2.dublicated_field having COUNT(*)>1
) t3
on t1.dublicated_field=t3.dublicated_field
and len(t1.field_kept)=t3.min_field_kept
微软有一个关于如何删除重复文件的非常简洁的指南。查看http://support.microsoft.com/kb/139444
简而言之，当你只有几行要删除时，下面是删除重复项的最简单方法:
SET rowcount 1;
DELETE FROM t1 WHERE myprimarykey=1;
Myprimarykey是行标识符。
我将rowcount设置为1，因为我只有两行复制。如果我复制了3行，那么我就会将rowcount设置为2，这样它就会删除它看到的前两行，只在表t1中留下一行。
