MySQL show sum of difference of two values

问题

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_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()

Which gives me below result

I want to sum up the kwh_diff and to show only one-row record not multiple like below

name customer_id msn sum_kwh_diff

Zeeshan 37010114711 4A60193390663 4.5

I have tried to do the following

 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.

回答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.

回答2:

You can’t use window functions within an aggregate function (while the opposite is possible), Here, you need to use a subquery, and aggregate in the outer query:

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

回答3:

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`

来源：https://stackoverflow.com/questions/62608035/mysql-show-sum-of-difference-of-two-values

标签

mysql

sql

select

sum

window-functions