MySQL show sum of difference of two values

Question

Below is my query.

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_dif         


        

Answer 1

You want to sum the differences between consecutive rows.
Say, for example, that you have these values for the column kwh:

kwh
---
10
12
14
17
25
32

so the differences are:

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

The sum of these differences is equal to 32-10 which is:

the diffference between the last value and the first value

So what you need is window function FIRST_VALUE() to obtain these values:

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()

and no subquery or aggregation is needed.

I kept in my code PARTITION BY n.customer_id because you use it in your code, although you may need PARTITION BY n.customer_id, m.msn.