Question

我有下表：

CREATE TABLE IF NOT EXISTS `Customers` (
    `id` INT AUTO_INCREMENT,
    `name` VARCHAR(20) NOT NULL,
    PRIMARY KEY(`id`)
);

CREATE TABLE IF NOT EXISTS `Orders` (
    `id` INT AUTO_INCREMENT,
    `id_cust` INT NOT NULL,
    `descr` VARCHAR(40),
    `price` INT NOT NULL,
    PRIMARY KEY(`id`),
    FOREIGN KEY(`id_cust`) REFERENCES `Customers`(`id`)
);

一个客户可以拥有多个订单。我想获得id_cust和支付最多（一个人）的订单总和。我的问题：

SELECT cust, max_orders_sum
FROM
(
    (
        SELECT MAX(orders_sum) AS max_orders_sum
        FROM (
            SELECT o.id_cust AS cust, SUM(o.price) AS orders_sum
            FROM Orders AS o
            GROUP BY o.id_cust
        ) AS same_query0
    ) AS step1
    INNER JOIN
    (
        SELECT o.id_cust AS cust, SUM(o.price) AS orders_sum 
        FROM Orders AS o
        GROUP BY o.id_cust
    ) AS same_query1
    ON (step1.max_orders_sum = same_query1.orders_sum)
);

主要问题： 如您所见，它具有相同的部分：same_query0和same_query1。有没有办法摆脱它们？或者，如果您知道实现目标的更好方式，请分享。我找到了一个简单的解决方案：

SELECT o.id_cust AS cust, SUM(o.price) AS orders_sum
FROM Orders AS o
GROUP BY o.id_cust
ORDER BY orders_sum DESC LIMIT 1;

但这不是解决问题的直接方法。

Answer 1

除非你创建一个像

这样的视图，否则我认为你不能比你已经做的更好。

create view v_cust_tot as
select  id_cust, sum(price) as cust_tot
from    Orders
group by id_cust

有了这个，你就可以像这样重写你的查询

select  id_cust, cust_tot
from    v_cust_tot
where   cust_tot = (select max(cust_tot) from v_cust_tot)

这只是在查询的紧凑性方面的改进，因为我认为性能与执行计划几乎完全相同

Answer 2

另一个不错的解决方案：

select id_cust, sum(price) from orders group by id_cust having sum(price) = 
(select max(prc) from
(select sum(price) as prc from orders group by id_cust) as tb);

改进获取具有最大值的行的查询

2 个答案: