select * from
(
select top 5 * from
( select sum(number) as amount,ISBN from tab_Order_detail group by ISBN )as tab order by amount desc) o
inner join
tab_bookinfo b
on o.ISBN=b.ISBN order by o.amount desc
( select sum(number) as amount,ISBN from tab_Order_detail group by ISBN )as tab order by amount desc) o
inner join
tab_bookinfo b
on o.ISBN=b.ISBN order by o.amount desc 做了一个inner join 内连接 得到一个表 (o inner join b on o.ISBN=b.ISBN)select top 5 * from 然后取前5行最后 select * 取全部结果
(
select top 5 * from
( select sum(number) as amount,ISBN from tab_Order_detail group by ISBN )as tab order by amount desc) o
inner join
tab_bookinfo b
on o.ISBN=b.ISBN order by o.amount desc
( select sum(number) as amount,ISBN from tab_Order_detail group by ISBN )as tab order by amount desc) o
inner join
tab_bookinfo b
on o.ISBN=b.ISBN order by o.amount desc 做了一个inner join 内连接 得到一个表 (o inner join b on o.ISBN=b.ISBN)select top 5 * from 然后取前5行最后 select * 取全部结果
解决方案 »
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我们又可令select2=select sum(number) as amount,ISBN from tab_Order_detail group by ISBN ,则selec1是从表selec2中查询前5行所得到的;则select1可被替换为select1=select top 5 * from select2 as tab order by amount desc,其中又通过as tab将select1命名为tab;select2就好理解了。仅做参考,错误之处敬请批评指正~~~
(select top 5 * from
( select sum(number) as amount,ISBN from tab_Order_detail
group by ISBN )as tab
order by amount desc
) o
inner join tab_bookinfo b
on o.ISBN=b.ISBN
order by o.amount desc
蓝色部分,以ISBN 分组,求number之和,然后按number之和降序取得前五条记录集。
然后和tab_bookinfo 连接,取出所有满足条件的纪录。
条件是 ISBN相等,并amount 降序排序