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都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？感谢。

Answer 1

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

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

这里我们使用subselect（来自同一result的前面时间戳的最大ip）获得前一个值。如果这是第一个值，IFNULL给出0，所以初始结果显示正确。

我还建议将以下索引添加到thetable：

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

Answer 2

试试这个 -

SET @f_rank = 0;
SET @s_rank = 0;
SET @f_date = NULL;
SET @f_ip = NULL;
SET @s_date = NULL;
SET @s_ip = NULL;
SELECT t1.date, t1.time, t1.ip, IF(t2.result IS NULL, t1.result, t1.result - t2.result) diff FROM
    (SELECT *, IF(@f_date = date AND @f_ip = ip, @f_rank:=@f_rank + 1, @f_rank:=1) AS rank, @f_date := date, @f_ip := ip FROM table1 ORDER BY date, ip, time) t1
  LEFT JOIN 
    (SELECT *, IF(@s_date = date AND @s_ip = ip, @s_rank:=@s_rank + 1, @s_rank:=1) AS rank, @s_date := date, @s_ip := ip FROM table1 ORDER BY date, ip, time) t2
  ON t1.date = t2.date AND t1.ip = t2.ip AND (t1.rank = t2.rank + 1);

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

MySQL - 找到同一个表的行之间的差异

2 个答案: