吨-SQL - 删除第二个值仅

Question

我想运行一个SQL语句，将只删除第二个值例如

delete from table1 where condition1 

我想这种说法只删除第二个值

如何我能完成这个吗？

我想澄清一下。我有一个名为field1的字段，它是一个自动编号，并且它是一个主键，它会递增。我想删除包含更多数量的

Answer 1

您也可以使用SQL服务器的ROW_NUMBER()函数对每一行进行编号，并根据您在内部查询（over (ORDER BY <myKey> asc)）中的自定义排序，使用此编号隔离第二个要删除的项目。这提供了很大的灵活性。

DELETE a FROM table1 
FROM table1 a 
JOIN (
     select ROW_NUMBER() over (ORDER BY <myKey> asc) as AutoNumber, <myKey> from table1 
) b on a.<myKey> = b.<myKey> 
WHERE condition1 
AND b.AutoNumber = 2 

Answer 2

试试这个记录：

DELETE MyTable 
FROM MyTable 
LEFT OUTER JOIN (
    SELECT MIN(id) as id, Col1, Col2, Col3 
    FROM MyTable 
    GROUP BY Col1, Col2, Col3 
) as KeepRows ON 
    MyTable.id= KeepRows.id 
WHERE 
    KeepRows.RowId IS NULL 

UPDATE

虽然这可能不会像“漂亮”为@杰弗里的工作原理。据我所知，@ Jeffrey's没有。请参阅下面的SQL（删除与SELECT *取代了演示）：

WITH TEMP as 
(
    SELECT 1 as id,'A' as a,'Z' as b 
    UNION 
    SELECT 2,'A','Z' 
    UNION 
    SELECT 3,'B','Z' 
    UNION 
    SELECT 4,'B','Z' 
) 

SELECT * 
FROM TEMP 
LEFT OUTER JOIN (
    SELECT MIN(id) as id, a, b 
    FROM TEMP 
    GROUP BY a, b 
) as KeepRows ON 
    temp.id= KeepRows.id 
WHERE 
    KeepRows.id IS NULL 

Answer 3

它已经有一段时间（所以我的语法我不太是正确的），而这可能不是最好的解决办法，但“学术”的答案会是这样的：

delete from table1 where condition1 
and field1 = (select max(field1) from table1 where condition1) 

Answer 4

是否只想删除最后一个副本，或者只删除第一个副本？

为所有，但第一种：（编辑使用CTE每@马丁的建议）

with target as (select * from table1 where condition1) 
delete from target goner 
where exists (select * from target keeper 
       where keeper.field1 < goner.field1) 

换句话说，如果以较低的field1的另一个匹配的记录，删除该记录。

编辑： 只删除最后：

with target as (select * from table1 where condition1) 
delete from target goner 
where exists (select * from target keeper 
       where keeper.field1 < goner.field1) 
    and not exists (select * from target missing 
        where missing.field1 > goner.field1) 

换句话说，如果以较低的field1的另一个匹配的记录，并有具有较高FIELD1没有匹配的记录，那么我们有最高的重复，所以核弹它。