题解 | #确定最佳顾客的另一种方式（二）#

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