我有下表:
++++++++++++++++++++++++++++++++++++++++++++++++++
| location | server | datetime | max_cpu |
++++++++++++++++++++++++++++++++++++++++++++++++++
| Chicago | 1 | 2013-05-01 00:00 | 10 |
| Chicago | 1 | 2013-05-01 01:00 | 15 |
| Chicago | 1 | 2013-05-01 02:00 | 11 |
| Chicago | 2 | 2013-05-01 00:00 | 8 |
| Chicago | 2 | 2013-05-01 01:00 | 12 |
| Chicago | 2 | 2013-05-01 02:00 | 13 |
| Atlanta | 1 | 2013-05-01 00:00 | 11 |
| Atlanta | 1 | 2013-05-01 01:00 | 12 |
| Atlanta | 1 | 2013-05-01 02:00 | 19 |
| Atlanta | 2 | 2013-05-01 00:00 | 21 |
| Atlanta | 2 | 2013-05-01 01:00 | 15 |
| Atlanta | 2 | 2013-05-01 02:00 | 17 |
我需要给定日期每个位置每个盒子的最大 CPU,例如
++++++++++++++++++++++++++++++++++++++++++++++++++
| location | server | datetime | max_cpu |
++++++++++++++++++++++++++++++++++++++++++++++++++
| Chicago | 1 | 2013-05-01 01:00 | 15 |
| Chicago | 2 | 2013-05-01 02:00 | 13 |
| Atlanta | 1 | 2013-05-01 02:00 | 19 |
| Atlanta | 2 | 2013-05-01 00:00 | 21 |
我知道如何针对单个条件(例如,仅位置)执行此操作并尝试扩展它(见下文),但它没有提供我需要的输出。
SELECT a.location, a.server, a.datetime, a.max_cpu
FROM mytable as a INNER JOIN
(
SELECT location, server, max(max_cpu) as max_cpu
FROM mytable
GROUP BY location, server
)
AS b ON
(
a.location = b.location
AND a.server = b.server
AND a.max_cpu = b.max_cpu
)
最佳答案
您可以通过找到最大 CPU 并连接回原始表来完成此操作。
看来你想要最大值的时间以及数量(这在文本中没有明确说明,但在结果中很清楚):
select t.*
from mytable t join
(select location, server, DATE(datetime) as thedate, MAX(max_cpu) as maxmaxcpu
from mytable t
group by location, server, DATE(datetime)
) lsd
on lsd.location = t.location and lsd.server = t.server and
lsd.thedate = DATE(t.datetime) and lsd.maxmaxcpu = t.max_cpu
这会计算每天的 maxcpu,然后返回以获取原始数据中的适当行 或行。如果有多个记录有 max,您将获得所有记录。如果您只需要一个,您可以将 group by location, server, day(datetime)
添加到查询中。
关于MySQL:为每个位置和服务器查找列的最大值,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/16319602/