题解 | #确定最佳顾客的另一种方式（二）#

select
    cust_name,
    t.total_price
from
    Customers c,
    Orders o,
    (
        select
            order_num,
            sum(item_price * quantity) as total_price
        from
            OrderItems
        group by
            order_num
    ) t
where
    t.order_num = o.order_num
    and o.cust_id = c.cust_id
having
    t.total_price >= 1000
order by
    t.total_price