MySQL显示两个值之差之和

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

我想总结一下kwh_diff并仅显示一行记录而不是多行记录，如下所示

name customer_id msn sum_kwh_diff

Zeeshan 37010114711 4A60193390663 4.5

我尝试过执行以下操作

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

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

您想要对连续行之间的差异求和。
例如，假设您有该列的这些值kwh:

kwh
---
10
12
14
17
25
32

所以差异是：

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

这些差异的总和等于32-10这是：

最后一个值和第一个值之间的差

所以你需要的是窗口函数FIRST_VALUE()获得这些值：

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。

我保留在我的代码中PARTITION BY n.customer_id因为你在代码中使用它，尽管你可能需要PARTITION BY n.customer_id, m.msn.