题解 | #确定最佳顾客的另一种方式（二）#

select
  cust_name,
  sum(quantity * item_price) total_price
from
  OrderItems a
  INNER JOIN Orders b on a.order_num = b.order_num
  INNER JOIN Customers c on b.cust_id = c.cust_id
group by
  cust_name
having
  total_price >= 1000
order by
  total_price;