题解 | #确定最佳顾客的另一种方式（二）#

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