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
试着使用:
SELECT linkorder
,Row_Number() OVER (
PARTITION BY linkorder ORDER BY linkorder DESC
) AS RowNum
FROM u_links
其他回答
-- 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
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。
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
试着使用:
SELECT linkorder
,Row_Number() OVER (
PARTITION BY linkorder ORDER BY linkorder DESC
) AS RowNum
FROM u_links
