Question

大家好我想查询当前日期减去上一个日期并使用SQL返回长度的变化（当前长度 - 前一个长度）。请参阅下表中当前减去前一长度

的变化样本

我无法想到查询如何做到这一点 SELECT日期，长度FROM length_data

date                 length          changes
-------------------  --------------  ------------
2000-08-29 10:30:00          147.98  147.98
2000-08-30 00:00:00          147.98  0
2000-09-02 10:30:00          156.51  8.53
2000-09-04 00:00:00          156.51  ....
2000-09-04 04:30:00          156.51  ....
2000-09-04 06:30:00          156.51  ....
2000-09-05 21:00:00          156.51
2000-09-06 03:00:00          156.51
2000-09-07 09:30:00          204.06
2000-09-07 10:30:00          204.06
2000-09-08 00:00:00          339.09
2000-09-08 12:30:00          395.78
2000-09-09 09:30:00          477.77
2000-09-10 02:30:00          737.77

Answer 1

您可以使用变量来模拟LAG窗口函数：

SELECT `date`, `length`,
       `length` - @prev AS changes,
       @prev := `length`
FROM length_data, (SELECT @prev := 0) AS a
ORDER BY `date`;

的 SqlFiddleDemo

我认为date是PRIMARY KEY/UNIQUE，否则它就不会稳定。

修改

要删除其他列，您可以使用子查询。

SELECT `date`, `length`, `changes` FROM (SELECT `date`, `length`, `length` - @prev AS changes, @prev := `length` FROM length_data,(SELECT @prev := 0) as a ORDER BY `date`) AS sub ORDER BY `date`;

的 SqlFiddleDemo2

输出：

╔══════════════════════════════╦═════════╦═════════╗ ║ date ║ length ║ changes ║ ╠══════════════════════════════╬═════════╬═════════╣ ║ August, 29 2000 10:30:00 ║ 147.98 ║ 147.98 ║ ║ August, 30 2000 00:00:00 ║ 147.98 ║ 0 ║ ║ September, 02 2000 10:30:00 ║ 156.51 ║ 8.53 ║ ║ September, 04 2000 00:00:00 ║ 156.51 ║ 0 ║ ║ September, 04 2000 04:30:00 ║ 156.51 ║ 0 ║ ║ September, 04 2000 06:30:00 ║ 156.51 ║ 0 ║ ║ September, 05 2000 21:00:00 ║ 156.51 ║ 0 ║ ║ September, 06 2000 03:00:00 ║ 156.51 ║ 0 ║ ║ September, 07 2000 09:30:00 ║ 204.06 ║ 47.55 ║ ║ September, 07 2000 10:30:00 ║ 204.06 ║ 0 ║ ║ September, 08 2000 00:00:00 ║ 339.09 ║ 135.03 ║ ║ September, 08 2000 12:30:00 ║ 395.78 ║ 56.69 ║ ║ September, 09 2000 09:30:00 ║ 477.77 ║ 81.99 ║ ║ September, 10 2000 02:30:00 ║ 737.77 ║ 260 ║ ╚══════════════════════════════╩═════════╩═════════╝

编辑2：

使用Jean Doux's idea仅使用一列来计算它：

SELECT `date`, length, CAST((-IF(@prev IS NULL,0,@prev) + (@prev := length)) AS DECIMAL(10,4)) AS changes FROM length_data,(SELECT @prev := NULL ) AS a ORDER BY `date`;

的 SqlFiddleDemo2

或最简单的方法是跳过应用程序层中的其他列。

Answer 2

SELECT
date, 
length,
@prevLength - (@prevLength:= length) as changes
FROM 
test, (SELECT @prevLength:= 0) as a
ORDER BY account_id ASC, value ASC

如何从减去日期获得结果

2 个答案: