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 AS c INNER JOIN Orders AS o
    ON c.cust_id = o.cust_id
    INNER JOIN OrderItems AS oi
    ON o.order_num = oi.order_num
GROUP BY
    c.cust_name
HAVING
    SUM(oi.item_price*oi.quantity) >= 1000
ORDER BY
    total_price