题解 | #确定最佳顾客的另一种方式（二）#

select
    cust_name,
    total_price
from
    Customers
    inner join Orders on Customers.cust_id = Orders.cust_id
    inner join (
        select
            order_num,
            sum(item_price * quantity) as total_price
        from
            OrderItems
        group by
            order_num
    ) as O on Orders.order_num = O.order_num
where
    total_price >= 1000
order by
    total_price