日期:2014-05-18  浏览次数:20459 次

【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 'Ma