题解 | 确定最佳顾客的另一种方式（二）

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