Question

我想计算连续行中两个值的差异。例如：计算前一天燃油价格的差异。如何使用SQL查询解决这个问题（值如下）。请帮助我

Date        Speed      Petrol      Diesel
2018-04-01  79.22       76.41       68.89   
2018-04-02  79.32       76.52       69.01   
2018-04-04  79.45       76.64       69.15   
2018-04-05  79.48       76.67       69.18   
2018-04-06  79.48       76.67       69.21   
2018-04-07  79.5        76.69       69.25   
2018-04-08  79.53       76.72       69.3    
2018-04-09  79.49       76.69       69.27   
2018-04-10  79.48       76.67       69.3    
2018-04-11  79.48       76.67       69.3    
2018-04-12  79.44       76.63       69.27   
2018-04-13  79.4        76.6        69.3

Answer 1

如果您没有支持窗口函数的版本，您可以使用子查询来获取昨天的价格

select `date`,speed,
          speed - ifnull((select speed from t t1 where t1.date < t.date order by t1.date desc limit 1),speed) yesterdayspeed,
          petrol,
          petrol - ifnull((select petrol from t t1 where t1.date < t.date order by t1.date desc limit 1),petrol) yesterdayspetrol,
          diesel,
          diesel - ifnull((select diesel from t t1 where t1.date < t.date order by t1.date desc limit 1),diesel) yesterdaysdiesel
from t;

+------------+-------+----------------+--------+------------------+--------+------------------+
| date       | speed | yesterdayspeed | petrol | yesterdayspetrol | diesel | yesterdaysdiesel |
+------------+-------+----------------+--------+------------------+--------+------------------+
| 2018-04-01 | 79.22 |           0.00 |  76.41 |             0.00 |  68.89 |             0.00 |
| 2018-04-02 | 79.32 |           0.10 |  76.52 |             0.11 |  69.01 |             0.12 |
| 2018-04-04 | 79.45 |           0.13 |  76.64 |             0.12 |  69.15 |             0.14 |
| 2018-04-05 | 79.48 |           0.03 |  76.67 |             0.03 |  69.18 |             0.03 |
| 2018-04-06 | 79.48 |           0.00 |  76.67 |             0.00 |  69.21 |             0.03 |
| 2018-04-07 | 79.50 |           0.02 |  76.69 |             0.02 |  69.25 |             0.04 |
| 2018-04-08 | 79.53 |           0.03 |  76.72 |             0.03 |  69.30 |             0.05 |
| 2018-04-09 | 79.49 |          -0.04 |  76.69 |            -0.03 |  69.27 |            -0.03 |
| 2018-04-10 | 79.48 |          -0.01 |  76.67 |            -0.02 |  69.30 |             0.03 |
| 2018-04-11 | 79.48 |           0.00 |  76.67 |             0.00 |  69.30 |             0.00 |
| 2018-04-12 | 79.44 |          -0.04 |  76.63 |            -0.04 |  69.27 |            -0.03 |
| 2018-04-13 | 79.40 |          -0.04 |  76.60 |            -0.03 |  69.30 |             0.03 |
+------------+-------+----------------+--------+------------------+--------+------------------+
12 rows in set (0.00 sec)

Answer 2

因为这对我来说似乎有些商业逻辑。最好在你的代码中执行此操作，而不是在我相信的sql查询中。

您可以获取行对象列表中的所有数据，但可以找到差异。

Answer 3

看起来是在MySQL 8.0中使用新窗口函数的好机会

假设其中一个与燃油价格有关，您希望按日期按升序排列结果，以及一个看起来像这样的表

CREATE TABLE tspeed (
  `id` INT auto_increment primary key,
  `Date` DATE NOT NULL,
  `Speed` DECIMAL(5,2),
  `Petrol` DECIMAL(5,2),
  `Diesel` DECIMAL(5,2)
);

您可以使用LAG（）函数来获取任何列的差异。

e.g。

SELECT `id`, `Date`, `Speed`,
  LAG(`Speed`) OVER w AS 'Prev Speed',
  `Speed` - LAG(`Speed`) OVER w AS 'Diff Speed',
  `Petrol`,
  LAG(`Petrol`) OVER w AS 'Prev Petrol',
  `Petrol` - LAG(`Petrol`) OVER w AS 'Diff Petrol',
  `Diesel`,
  LAG(`Diesel`) OVER w AS 'Prev Diesel',
   `Diesel` - LAG(`Diesel`) OVER w AS 'Diff Diesel'
FROM tspeed
WINDOW w AS (ORDER BY `Date` ASC);

https://www.db-fiddle.com/f/nLTRhsUKELy6YpMSAEKrCN/5

如何计算连续行中的值的差异？

3 个答案: