How to get the difference between consecutive rows in MySQL? - mysql

Question

I have a table in mysql database this data.
id date number qty
114 07-10-2018 200 5
120 01-12-2018 300 10
123 03-02-2019 700 12
1126 07-03-2019 1000 15
I want to calculate difference between two consecutive rows and i need output format be like:
id date number diff qty avg
114 07-10-2018 200 0 5 0
120 01-12-2018 300 100 10 10
123 03-02-2019 700 400 12 33.33
1126 07-03-2019 1000 300 15 20
Any one know how to do this in mysql query? I want first value of diff and avg column to be 0 and rest is the difference.

Answer 1

For MySQL 8 then use Lag window function.
SELECT
test.id,
test.date,
test.number,
test.qty,
IFNULL(test.number - LAG(test.number) OVER w, 0) AS diff,
ROUND(IFNULL(test.number - LAG(test.number) OVER w, 0)/ test.qty, 2) AS 'Avg'
FROM purchases test
WINDOW w AS (ORDER BY test.`date` ASC);
For MySQL 5.7 or lesser version
We can use the MySQL variable to do this job. Consider your table name is test.
SELECT
test.id,
test.date,
test.number,
test.qty,
#diff:= IF(#prev_number = 0, 0, test.number - #prev_number) AS diff,
ROUND(#diff / qty, 2) 'avg',
#prev_number:= test.number as dummy
FROM
test,
(SELECT #prev_number:= 0 AS num) AS b
ORDER BY test.`date` ASC;
-------------------------------------------------------------------------------
Output:
| id | date | number| qty | diff | avg | dummy |
-----------------------------------------------------------------
| 114 | 2018-10-07 | 200 | 5 | 0 | 0.00 | 200 |
| 120 | 2018-12-01 | 300 | 10 | 100 | 10.00 | 300 |
| 123 | 2019-02-03 | 700 | 12 | 400 | 33.33 | 700 |
| 1126 | 2019-03-07 | 1000 | 15 | 300 | 20.00 | 1000 |
Explaination:
(SELECT #prev_number:= 0 AS num) AS b
we initialized variable #prev_number to zero in FROM clause and joined with each row of the test table.
#diff:= IF(#prev_number = 0, 0, test.number - #prev_number) AS diff First we are generating difference and then created another variable diff to reuse it for average calculation. Also we included one condition to make the diff for first row as zero.
#prev_number:= test.number as dummy we are setting current number to this variable, so it can be used by next row.
Note: We have to use this variable first, in both difference as well as average and then set to the new value, so next row can access value from the previous row.
You can skip/modify order by clause as per your requirements.

Answer 2

There could be better ways to do this, but try this:
SELECT A.id,
A.date,
A.number,
A.qty,
A.diff,
B.avg
FROM
(SELECT *, abs(LAG(number, 1, number) OVER (ORDER BY id) - number) AS 'diff'
FROM table) AS A
JOIN
(SELECT *, abs(LAG(number, 1, number) OVER (ORDER BY id) - number)/qty AS 'avg' FROM table) AS B
ON A.id = B.id;