【关于按小时统计SQL】请各位帮帮忙,该怎么处理

【关于按小时统计SQL】请各位帮帮忙
表结构如下：

MID[设备编号] AA[浮点型] BB[某浮点型] DATE[时间]
Machine1 5.5 6.5 2012-4-20 12:05:00
Machine1 5.6 6.6 2012-4-20 12:08:00
Machine1 6.7 2.3 2012-4-20 01:10:00
Machine1 99.99 98.88 2012-4-20 01:59:00
Machine1 9.0 3.3 2012-4-20 02:15:00
Machine1 7.2 6.8 2012-4-20 03:25:00
Machine1 5.6 9.8 2012-4-20 04:30:00
Machine1 4.2 4.4 2012-4-20 05:35:00
Machine1 8.8 3.6 2012-4-20 06:40:00

表说明：
MID指的是一个设备编号
每隔几分钟就插入表中一些数据

要求查询结果

字段 0-1小时内 1-2小时内 2-3小时内 3-4小时内 .... 22-23小时内
AA 5.6 99.99 9.0 5.6
BB 6.6 98.88 3.3 9.8 .... ....

某个小时内可能存在多条记录，只要查询最新日期的那一条就行

------解决方案--------------------
楼主，请再详细描述一下你的结果，比如 AA，在“0-1小时内”的值为什么是5.6，而在“1-2小时内”的值是99.99

------解决方案--------------------

SQL code

---------------------------------
--  Author: HEROWANG(让你望见影子的墙)
--  Date  : 2012-04-21 07:15:18
--  blog  : blog.csdn.net/herowang
---------------------------------
 
IF OBJECT_ID('[tb]') IS NOT NULL 
    DROP TABLE [tb]
go
CREATE TABLE [tb] (MID VARCHAR(8),AA NUMERIC(4,2),BB NUMERIC(4,2),DATE DATETIME)
INSERT INTO [tb]
SELECT 'Machine1',5.5,6.5,'2012-4-20 00:05:00' UNION ALL
SELECT 'Machine1',5.6,6.6,'2012-4-20 00:08:00' UNION ALL
SELECT 'Machine1',6.7,2.3,'2012-4-20 01:10:00' UNION ALL
SELECT 'Machine1',99.99,98.88,'2012-4-20 01:59:00' UNION ALL
SELECT 'Machine1',9.0,3.3,'2012-4-20 02:15:00' UNION ALL
SELECT 'Machine1',7.2,6.8,'2012-4-20 03:25:00' UNION ALL
SELECT 'Machine1',5.6,9.8,'2012-4-20 04:30:00' UNION ALL
SELECT 'Machine1',4.2,4.4,'2012-4-20 05:35:00' UNION ALL
SELECT 'Machine1',8.8,3.6,'2012-4-20 06:40:00'

--select *,convert(char(13),date,120)  from [tb]
go
with cte
as(select row=row_number() over(partition by mid,convert(char(13),date,120) order by date desc),* from tb),
cte2 as(select row, mid,ziduan='AA',AA,date from cte where row=1 union select row, mid,'BB',BB,date from cte where row=1 )

select ziduan,
[0-1]=max(case when ziduan='AA' and date >= '2012-04-20 00:00:00' and date <'2012-04-20 01:00:00' then AA 
          when ziduan='BB' and date >= '2012-04-20 00:00:00' and date <'2012-04-20 01:00:00' then AA 
          end),
[1-2]=max(case when ziduan='AA' and date >= '2012-04-20 01:00:00' and date <'2012-04-20 02:00:00' then AA 
          when ziduan='BB' and date >= '2012-04-20 01:00:00' and date <'2012-04-20 02:00:00' then AA 
          end),
[2-3]=max(case when ziduan='AA' and date >= '2012-04-20 02:00:00' and date <'2012-04-20 03:00:00' then AA 
          when ziduan='BB' and date >= '2012-04-20 02:00:00' and date <'2012-04-20 03:00:00' then AA 
          end)
…… 剩下的一次类推
from cte2
group by ziduan

楼主的工作就是构造'2012-04-20 02:00:00' 时间的起始点，这里用的是固定的日期，

------解决方案--------------------

借用一下 HEROWANG 的测试数据。

SQL code

IF OBJECT_ID('[tb]') IS NOT NULL 
    DROP TABLE [tb]
go
CREATE TABLE [tb] (MID VARCHAR(8),AA NUMERIC(4,2),BB NUMERIC(4,2),DATE DATETIME)
INSERT INTO [tb]
SELECT 'Machine1',5.5,6.5,'2012-4-20 00:05:00' UNION ALL
SELECT 'Machine1',5.6,6.6,'2012-4-20 00:08:00' UNION ALL
SELECT 'Machine1',6.7,2.3,'2012-4-20 01:10:00' UNION ALL
SELECT 'Machine1',99.99,98.88,'2012-4-20 01:59:00' UNION ALL
SELECT 'Machine1',9.0,3.3,'2012-4-20 02:15:00' UNION ALL
SELECT 'Machine1',7.2,6.8,'2012-4-20 03:25:00' UNION ALL
SELECT 'Machine1',5.6,9.8,'2012-4-20 04:30:00' UNION ALL
SELECT 'Machine1',4.2,4.4,'2012-4-20 05:35:00' UNION ALL
SELECT 'Machine1',8.8,3.6,'2012-4-20 06:40:00';

--select *, convert(varchar(13), date, 120) as hourP from tb;
with T as
(
    select b.MID, left(a.hourP, 10) as Days, cast(right(a.hourP, 2) as int) as hour, b.AA, b.BB
    from (
            select  convert(varchar(13), date, 120) as hourP, max(date) as time
            from tb
            group by convert(varchar(13), date, 120)
          ) a
    join tb b on a.time = b.date
)
select 'AA' as 字段, * 
from (select mid, days, hour, AA from T ) a
pivot
(
    max(a.AA) for a.hour
    in([0], [1], [2], [3], [4], [5], [6], [7], [8], [9], [10], [11],
       [12], [13], [14], [15], [16], [17], [18], [19], [20], [21], [22], [23])
) b
union all
select 'BB' as 字段, * 
from (select mid, days, hour, BB from T ) a
pivot
(
    max(a.BB) for a.hour
    in([0], [1], [2], [3], [4], [5], [6], [7], [8], [9], [10], [11],
       [12], [13], [14], [15], [16], [17], [18], [19], [20], [21], [22], [23])
) b

/*
字段   mid      days          0       1       2       3       4       5       6       7       8       9       10      11      12      13      14      15      16      17      18      19      20      21      22      23
---- -------- ------------- --------------------------------------- --------------------------------------- --------------------------------------- --------------------------------------- --------------------------------
AA   Machine1 2012-04-20    5.60    99.99   9.00    7.20    5.60    4.20    8.80    NULL    NULL    NULL    NULL    NULL    NULL    NULL    NULL    NULL    NULL    NULL    NULL    NULL    NULL    NULL    NULL    NULL
BB   Machine1 2012-04-20    6.60    98.88   3.30    6.80    9.80    4.40    3.60    NULL    NULL    NULL    NULL    NULL    NULL    NULL    NULL    NULL    NULL    NULL    NULL    NULL    NULL    NULL    NULL    NULL

(2 行受影响)


*/

【关于按小时统计SQL】请各位帮帮忙,该怎么处理

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