题解 | 确定最佳顾客的另一种方式（二）

select
    cust_name,
    sum(total_price) total_price
from
    (
        select
            cust_name,
            item_price * quantity total_price
        from
            OrderItems ori
            join Orders ord on ori.order_num = ord.order_num
            join Customers cus on ord.cust_id = cus.cust_id
    ) a
group by
    cust_name
having
    total_price >= 1000
order by
    total_price