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
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。
其他回答
试着使用:
SELECT linkorder
,Row_Number() OVER (
PARTITION BY linkorder ORDER BY linkorder DESC
) AS RowNum
FROM u_links
从一个巨大的(几百万条记录)表中删除重复项可能需要很长时间。我建议将所选行的批量插入到临时表中，而不是删除。
--REWRITING YOUR CODE(TAKE NOTE OF THE 3RD LINE) WITH CTE AS(SELECT NAME,ROW_NUMBER()
OVER (PARTITION BY NAME ORDER BY NAME) ID FROM @TB) SELECT * INTO #unique_records FROM
CTE WHERE ID =1;
DELETE from search
where id not in (
select min(id) from search
group by url
having count(*)=1
union
SELECT min(id) FROM search
group by url
having count(*) > 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
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;
