MySQL - 找到同一个表的行之间的差异

Question

我有一个包含带时间戳的聚合结果的表 - 这意味着每次每个结果是到目前为止的总数:

date       | time  | ip       | result
---------------------------------------
2011-03-01 | 10:00 | 10.0.0.1 | 200
2011-03-01 | 11:00 | 10.0.0.1 | 303
2011-03-01 | 12:00 | 10.0.0.1 | 415
2011-03-01 | 13:00 | 10.0.0.1 | 628
2011-03-01 | 10:00 | 10.0.0.2 | 198
2011-03-01 | 11:00 | 10.0.0.2 | 234
2011-03-01 | 12:00 | 10.0.0.2 | 373
2011-03-01 | 13:00 | 10.0.0.2 | 512

我正在尝试制定一个查询,它将获得每个时间范围之间的增量:

date       | time  | ip       | diff
---------------------------------------
2011-03-01 | 10:00 | 10.0.0.1 | 200
2011-03-01 | 11:00 | 10.0.0.1 | 103
2011-03-01 | 12:00 | 10.0.0.1 | 112
2011-03-01 | 13:00 | 10.0.0.1 | 213
2011-03-01 | 10:00 | 10.0.0.2 | 198
2011-03-01 | 11:00 | 10.0.0.2 |  36
2011-03-01 | 12:00 | 10.0.0.2 | 139
2011-03-01 | 13:00 | 10.0.0.2 | 139
...

因此每个日期/ ip分组的每一行都会减去它之前的那一行(或0).有什么简单的方法吗？谢谢.

Answer 1

这是一个没有变量的解决方案.我假设你在一个名为的表中有你的initail数据thetable.

SELECT date, time, ip,
    result - IFNULL( (
        SELECT MAX( result ) 
        FROM thetable
        WHERE ip = t1.ip
        AND ( date < t1.date
            OR date = t1.date AND time < t1.time )
    ) , 0) AS diff
FROM thetable AS t1
ORDER BY ip, date, time

这里我们通过子选择获得前一个值(result前一个时间戳的最大值ip).IFNULL如果这是第一个值,则给出0,因此初始结果显示正确.

我还建议将以下索引添加到thetable:

CREATE INDEX sort1 ON thetable (ip, date, time);