刚发的帖子结贴了...
然后自己看了下回帖
select t.* from a t where y = (select max(y) from a where x = t.x)select t.* from a t where not exists (select 1 from a where x = t.x and y > t.y)
知道可以实现
但不是很清楚原理...
求高手解释下....
先谢啦...
然后自己看了下回帖
select t.* from a t where y = (select max(y) from a where x = t.x)select t.* from a t where not exists (select 1 from a where x = t.x and y > t.y)
知道可以实现
但不是很清楚原理...
求高手解释下....
先谢啦...
使用 EXISTS 和 NOT EXISTS 引入的子查询可用于两种集合原理的操作:交集与差集。两个集合的交集包含同时属于两个原集合的所有元素。差集包含只属于两个集合中的第一个集合的元素。city 列中 authors 和 publishers 的交集是作者和出版商共同居住的城市的集合。USE pubs
SELECT DISTINCT city
FROM authors
WHERE EXISTS
(SELECT *
FROM publishers
WHERE authors.city = publishers.city)下面是结果集:city
--------
Berkeley(1 row(s) affected)当然,该查询可以写成一个简单的联接。USE pubs
SELECT DISTINCT authors.city
FROM authors INNER JOIN publishers
ON authors.city = publishers.citycity 列中 authors 和 publishers 的差集是作者所居住的、但没有出版商居住的所有城市的集合,也就是除 Berkeley 以外的所有城市。USE pubs
SELECT DISTINCT city
FROM authors
WHERE NOT EXISTS
(SELECT *
FROM publishers
WHERE authors.city = publishers.city)该查询也可以写成:USE pubs
SELECT DISTINCT city
FROM authors
WHERE city NOT IN
(SELECT city
FROM publishers)©1988-2000 Microsoft Corporation。保留所有权利。
select me.* from a me where not exists (select 1 from a where x = me.x and y > me.y)
sql比较菜,各位见谅....
特此感谢'爱新觉罗.毓华'