MySQL-查找同一表的行之间的差异
我有一个表,其中包含带有时间戳的聚合结果,这意味着每次的结果都是迄今为止的总和:MySQL-查找同一表的行之间的差异,mysql,query-help,Mysql,Query Help,我有一个表,其中包含带有时间戳的聚合结果,这意味着每次的结果都是迄今为止的总和: 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 | 1
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,该查询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分组的每一行减去它前面的一行(或0)。
有什么简单的方法吗?谢谢。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 |
+------------+----------+----------+------+
这是一个没有变量的解决方案。我假设您在一个名为
thetable
的表中有初始数据
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
您可能需要详细说明如何处理日期更改。结果是每天从零开始,还是仅在文件开始时才从零开始?如果有11:00的数据点,而当天之前没有10:00、只有09:00或没有数据点,该怎么办?很好的解决方案,效率更高。谢谢。它在处理
date
字段方面有点问题。现在没事了。我也遇到了一些问题,变量;)
CREATE INDEX sort1 ON thetable (ip, date, time);