题解 | #确定最佳顾客的另一种方式（二）#

select cust_name,total_price
from 
(select
    order_num,
    sum(item_price * quantity) as total_price
from
    OrderItems
group by
    order_num
having total_price>=1000) as s1 inner join 
(select order_num,cust_name from Orders as o inner join Customers as c on o.cust_id =c.cust_id) as s2 
on s1.order_num=s2.order_num order by total_price