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.
最佳答案
对于 MySQL 8 然后使用 Lag窗函数。
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);
适用于 MySQL 5.7 或更低版本
我们可以使用 MySQL variable做这个工作。假设您的表名称是 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 |
解释:
(SELECT @prev_number:= 0 AS num) AS b
我们在 FROM 子句中将变量 @prev_number 初始化为零,并与 test 表的每一行连接。@diff:= IF(@prev_number = 0, 0, test.number - @prev_number) AS diff
首先我们生成差异然后创建另一个变量diff重用它进行平均计算。我们还包括一个条件,使第一行的差异为零。@prev_number:= test.number as dummy
我们将当前的number 设置为此变量,以便下一行可以使用它。
注意:我们必须首先使用这个变量,在difference和average中,然后设置为新值,所以下一行可以访问上一行的值。
您可以根据需要跳过/修改order by
子句。
关于mysql - 如何获得MySQL中连续行之间的差异?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/57613990/