如何获取MySQL中连续行之间的差异？

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;