MySQL显示两个值的差异之和答案

【问题标题】：MySQL show sum of difference of two valuesMySQL显示两个值的差异之和
【发布时间】：2020-10-17 20:09:46
【问题描述】：

以下是我的查询。

SELECT n.`name`,n.`customer_id`,m.`msn`, m.kwh,
   m.kwh - LAG(m.kwh) OVER(PARTITION BY n.`customer_id` ORDER BY m.`data_date_time`) AS kwh_diff
FROM mdc_node n
INNER JOIN `mdc_meters_data` m ON n.`customer_id` = m.`cust_id`
WHERE n.`lft` = 5 AND n.`icon` NOT IN ('folder')
AND m.`data_date_time` BETWEEN NOW() - INTERVAL 30 DAY AND NOW()

这给了我下面的结果

我想总结 kwh_diff 并只显示一行记录而不是如下所示的多个记录

name customer_id msn sum_kwh_diff

Zeeshan 37010114711 4A60193390663 4.5

我尝试过以下操作

 SUM(m.kwh - LAG(m.kwh) OVER(PARTITION BY n.`customer_id` ORDER BY m.`data_date_time`)) AS sum_kwh_diff

得到Error Code: 4074 Window functions can not be used as arguments to group functions.

【问题讨论】：

标签： mysql sql select sum window-functions

【解决方案1】：

您不能在聚合函数中使用窗口函数（反之亦然），在这里，您需要使用子查询，并在外部查询中聚合：

SELECT name, customer_id, SUM(kwh_diff) sum_kwh_diff
FROM (
    SELECT n.`name`,n.`customer_id`,m.`msn`, m.kwh,
       m.kwh - LAG(m.kwh) OVER(PARTITION BY n.`customer_id` ORDER BY m.`data_date_time`) AS kwh_diff
    FROM mdc_node n
    INNER JOIN `mdc_meters_data` m ON n.`customer_id` = m.`cust_id`
    WHERE n.`lft` = 5 AND n.`icon` NOT IN ('folder')
    AND m.`data_date_time` BETWEEN NOW() - INTERVAL 30 DAY AND NOW()
) t
GROUP BY name, customer_id

【讨论】：

【解决方案2】：

做一个外部查询

SELECT
`name`,`customer_id`,`msn`, SUM(kwh_diff) kwh_diff
FROM
(
    SELECT n.`name`,n.`customer_id`,m.`msn`, m.kwh,
       m.kwh - LAG(m.kwh) OVER(PARTITION BY n.`customer_id` ORDER BY m.`data_date_time`) AS kwh_diff
    FROM mdc_node n
    INNER JOIN `mdc_meters_data` m ON n.`customer_id` = m.`cust_id`
    WHERE n.`lft` = 5 AND n.`icon` NOT IN ('folder')
    AND m.`data_date_time` BETWEEN NOW() - INTERVAL 30 DAY AND NOW() ) t1
GROUP BY `name`,`customer_id`,`msn`

【讨论】：

【解决方案3】：

您想对连续行之间的差异求和。
例如，假设您在列 kwh 中有这些值：

kwh
---
10
12
14
17
25
32

所以区别是：

kwh_diff
--------
0
12-10
14-12
17-14
25-17
32-25

这些差异的总和等于32-10，即：

最后一个值和第一个值的区别

所以你需要的是窗口函数FIRST_VALUE() 来获取这些值：

SELECT DISTINCT n.`name`, n.`customer_id`, m.`msn`, 
   FIRST_VALUE(m.kwh) OVER (PARTITION BY n.`customer_id` ORDER BY m.`data_date_time` DESC) -
   FIRST_VALUE(m.kwh) OVER (PARTITION BY n.`customer_id` ORDER BY m.`data_date_time` ASC) AS kwh_diff
FROM mdc_node n
INNER JOIN `mdc_meters_data` m ON n.`customer_id` = m.`cust_id`
WHERE n.`lft` = 5 AND n.`icon` NOT IN ('folder')
AND m.`data_date_time` BETWEEN NOW() - INTERVAL 30 DAY AND NOW()

并且不需要子查询或聚合。

我保留了我的代码PARTITION BY n.customer_id，因为您在代码中使用了它，尽管您可能需要PARTITION BY n.customer_id, m.msn。

【讨论】：