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

MAke an outer Query

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`