题解 | 确定最佳顾客的另一种方式（二）

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

如果有两张以上的多表查询可以使用：

select....

from....

where.....；

这样的结构，可以少写两个inner join。