分组和汇总列

问题描述

| 根据下表，（在两种服务器风格中）最佳方法是什么： 将在一分钟/小时/天之内的所有行分组，并获得最大列\'CounterC \'吗？ 示例：在\'now \'和\'now \'之间-1天，每小时可获得Max（CounterC）。 示例2：在\'now \'和\'now \'之间-30天之间，每天获取Max（CounterC）。 显然，行必须分组，但是如何？ MS SQL

CREATE TABLE [dbo].[DE0000000D102D1D](
[index] [bigint] IDENTITY(1,1) NOT NULL,[TimeStamp] [datetime] NOT NULL,[CounterA] [bigint] NOT NULL,[CounterB] [bigint] NOT NULL,[CounterC] [bigint] NOT NULL,[CounterD] [bigint] NOT NULL,)

的MySQL

CREATE TABLE `de0000000d102d1d` (
 `index` bigint(20) NOT NULL AUTO_INCREMENT,`TimeStamp` datetime NOT NULL,`CounterA` bigint(20) NOT NULL,`CounterB` bigint(20) NOT NULL,`CounterC` bigint(20) NOT NULL,`CounterD` bigint(20) NOT NULL,PRIMARY KEY (`index`)
)

一些示例数据：

index  TimeStamp                CounterA  CounterB  CounterC   CounterD
-----  -----------------------  --------  --------  ---------  --------
1      2011-03-07 14:25:32.000  0         1         347406352  916
2      2011-03-07 14:26:32.000  0         1         347407169  916
3      2011-03-07 14:27:32.000  0         1         347407978  916
4      2011-03-07 14:28:31.000  0         1         347408617  916
5      2011-03-07 14:29:31.000  0         1         347409087  916
6      2011-03-07 14:30:30.000  0         1         347409557  916
7      2011-03-07 14:31:09.000  0         1         347409845  916

提前致谢！ 编辑：实际上是每个间隔我想要的Max（CounterC），而不是总和。     

解决方法

        对于SQL Server

-- Last 30 days grouped by day
select dateadd(day,datediff(day,D.[TimeStamp]),0) as [day],max(D.CounterC) as MaxC
from DE0000000D102D1D as D
where D.[TimeStamp] between dateadd(d,-30,getdate()) and getdate()
group by dateadd(day,0)

-- Last day grouped by the hour
select dateadd(hour,datediff(hour,0) as [Hour],-1,getdate()) and getdate()
group by dateadd(hour,0)

    ,        对于MySQL，这是相同的查询：

-- Last 30 days grouped by day
select date_format( timestamp,\'%Y:%m:%d\' ) `day`,max(D.CounterC) as MaxC
from `DE0000000D102D1D` as D
where D.`TimeStamp` between timestampadd(day,now()) and now()
group by date_format( timestamp,\'%Y:%m:%d\' )
order by `day` ASC;

-- Last day grouped by the hour
select date_format( timestamp,\'%Y:%m:%d %H\' ) as `Hour`,\'%Y:%m:%d %H\' )
order by `Hour` ASC;