题解 | #确定最佳顾客的另一种方式（二）#

select c.cust_name,sum(item_price*quantity) total_price
from Orders o
join Customers c
on c.cust_id =o.cust_id
join OrderItems oi
on o.order_num =oi.order_num
group by c.cust_name
having total_price>=1000
order by total_price;