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 Customers c,OrderItems oi,Orders o
# where  c.cust_id=o.cust_id and oi.order_num=o.order_num
# group by c.cust_name
# having total_price>=1000
# order by total_price

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