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 Orders as o
inner join OrderItems as oi
on o.order_num = oi.order_num

inner join Customers as c
on o.cust_id = c.cust_id

group by cust_name
having total_price>=1000
order by total_price