我有以下表格：
Date Reading1 Reading2
2017-02-15 07:00:00.0000000 33 30
2017-02-15 07:15:00.0000000 32 31
2017-02-15 07:30:00.0000000 32 31
2017-02-15 07:45:00.0000000 33 30
2017-02-15 08:00:00.0000000 33 28
2017-02-15 08:15:00.0000000 32 29
2017-02-15 08:30:00.0000000 32 31
2017-02-15 08:45:00.0000000 34 31
2017-02-15 09:00:00.0000000 34 31
2017-02-15 09:15:00.0000000 34 30
2017-02-15 09:30:00.0000000 31 30
2017-02-15 09:45:00.0000000 32 32
........
2017-02-16 06:15:00.0000000 32 31
2017-02-16 06:30:00.0000000 35 32
2017-02-16 06:45:00.0000000 34 30
2017-02-16 07:00:00.0000000 34 31
我可以根据小时或日期对Reading1和Reading2列进行求和，但我的问题是我想在第二天早上7点到晚上7点，然后是第二天晚上7点到早上7点之间总结一下。任何帮助都会非常值得赞赏。
＆＃39;对于小时的总和，我使用以下查询＆＃39;
- 按小时计算
select datepart(hour,Date), SUM(Reading1), SUM(Reading2)
from #LocalTempTable
group by
datepart(hour,Date),
dateadd(d, 0, datediff(d, 0,Date))
答案 0 :(得分：0)
假设您的[日期]列是DATETIME列，您可以这样做:(基本上它的作用是将时间范围从上午7点到下午7点划分为一个，将7 PM-7AM划分为另一个。
select FORMAT(dateadd(hour,-7,[date]), 'yyyy-MM-dd') + case when DATEPART(hour,dateadd(hour,-7,[date])) between 0 and 11 then ' 7AM-7PM' ELSE ' 7PM-7AM' END as [TimeRange], SUM(Reading1), SUM(Reading2)
from #LocalTempTable
group by FORMAT(dateadd(hour,-7,[date]), 'yyyy-MM-dd') + case when DATEPART(hour,dateadd(hour,-7,[date])) between 0 and 11 then ' 7AM-7PM' ELSE ' 7PM-7AM' END
答案 1 :(得分：0)
假设：
我们需要每天汇总数据（不计算所有天的总和）
我们认为准确度为分钟，因此7AM = 420分钟（从0:00 AM开始）和7PM = 1140分钟
我们将一天分成2组：第1组＆gt;今天上午7点和＆lt;今天晚上7点，第2组＆gt; =今天晚上7点和明天＆lt; = 7 AM（E.G 20170228将有2组： 20170228_1和20170228_2）
然后你可以用这个：
SELECT
CASE WHEN DATEPART(hh, date)*60 + DATEPART(mi, date) <= 420
THEN CONVERT(char(8), date - 1, 112) + '_2'
WHEN DATEPART(hh, date)*60 + DATEPART(mi, date) >= 1140
THEN CONVERT(char(8), date, 112) + '_2'
ELSE CONVERT(char(8), date, 112) + '_1'
END AS date_group,
SUM(reading1),
SUM(reading2)
FROM table_name
GROUP BY
CASE WHEN DATEPART(hh, date)*60 + DATEPART(mi, date) <= 420
THEN CONVERT(char(8), date - 1, 112) + '_2'
WHEN DATEPART(hh, date)*60 + DATEPART(mi, date) >= 1140
THEN CONVERT(char(8), date, 112) + '_2'
ELSE CONVERT(char(8), date, 112) + '_1'
END;
答案 2 :(得分：0)
您可以使用案例陈述对小时进行分组以获得所需的结果：
select year(Date) as DateYear,
datepart(dy, Date) as DayOfYear,
case when datepart(hour,Date) >= 7 and datepart(hour, Date) < 19 then '7AM - 7PM' else '7PM - 7AM' end as HourGroup,
SUM(Reading1),
SUM(Reading2)
from #LocalTempTable
group by year(Date) as DateYear,
datepart(dy, Date) as DayOfYear,
case when datepart(hour,Date) >= 7 and datepart(hour, Date) < 19 then '7AM - 7PM' else '7PM - 7AM' end
答案 3 :(得分：0)
这是一个包含一些示例数据的完整脚本：
CREATE TABLE #Readings ([Date] DateTime, Reading1 int, Reading2 int)
INSERT INTO #Readings ([Date], Reading1, Reading2) VALUES ('2017-02-15 06:45:00', 1, 1)
INSERT INTO #Readings ([Date], Reading1, Reading2) VALUES ('2017-02-15 07:00:00', 2, 2)
INSERT INTO #Readings ([Date], Reading1, Reading2) VALUES ('2017-02-15 07:15:00', 3, 3)
INSERT INTO #Readings ([Date], Reading1, Reading2) VALUES ('2017-02-15 18:45:00', 4, 4)
INSERT INTO #Readings ([Date], Reading1, Reading2) VALUES ('2017-02-15 19:15:00', 5, 5)
INSERT INTO #Readings ([Date], Reading1, Reading2) VALUES ('2017-02-16 06:45:00', 6, 6)
INSERT INTO #Readings ([Date], Reading1, Reading2) VALUES ('2017-02-16 07:00:00', 7, 7)
SELECT
DATEADD(hh, (Half * 12) + 7, ModifiedDay) AS StartPeriod,
SUM(Reading1) AS SumOfReading1,
SUM(Reading2) AS SumOfReading2
FROM (
SELECT
[Date],
CAST(CAST(ModifiedDate AS DATE) AS DATETIME) AS [ModifiedDay],
DATEPART(hh, ModifiedDate) AS [DatePart],
CASE WHEN DATEPART(hh, ModifiedDate) < 12 THEN 0 ELSE 1 END AS Half,
Reading1,
Reading2
FROM (
SELECT
[Date],
DATEADD(hh, -7, [Date]) AS [ModifiedDate],
Reading1,
Reading2
FROM #Readings
) t
) t
GROUP BY DATEADD(hh, (Half * 12) + 7, ModifiedDay), Half
ORDER BY 1, 2
DROP TABLE #Readings
StartPeriod SumOfReading1 SumOfReading2
2017-02-14 19:00:00.000 1 1
2017-02-15 07:00:00.000 9 9
2017-02-15 19:00:00.000 11 11
2017-02-16 07:00:00.000 7 7
答案 4 :(得分：0)
不使用日历表或cte：
测试设置：http://rextester.com/ZENTG4450
select
FromDate = convert(varchar(10)
,min(dateadd(day,(datediff(hour,0,date)-7)/24,0))
,120)
, ThruDate = convert(varchar(10)
,max(dateadd(day,(datediff(hour,0,date)+5)/24,0))
,120)
, Hours = case ((datediff(hour,0,date)+5)/12)%2
when 1
then '7 AM to 7 PM'
else '7 PM to 7 AM'
end
, SumReading1=sum(Reading1)
, SumReading2=sum(Reading2)
from t
group by (datediff(hour,0,date)+5)/12
返回：
+------------+------------+--------------+-------------+-------------+
| FromDate | ThruDate | Hours | SumReading1 | SumReading2 |
+------------+------------+--------------+-------------+-------------+
| 2017-02-15 | 2017-02-15 | 7 AM to 7 PM | 392 | 364 |
| 2017-02-15 | 2017-02-16 | 7 PM to 7 AM | 101 | 93 |
| 2017-02-16 | 2017-02-16 | 7 AM to 7 PM | 34 | 31 |
+------------+------------+--------------+-------------+-------------+
使用日历表或cte：
测试设置：http://rextester.com/QOC88855
declare @fromdate date = '20170201'
declare @thrudate date = '20170228'
;with n as (select n from (values(0),(1),(2),(3),(4),(5),(6),(7),(8),(9)) t(n))
, dates as (
select top ((datediff(day, @fromdate, @thrudate)+1)*2)
[FromDate]=dateadd(hour,7+12*((row_number() over (order by (select 1)) -1)%2)
,convert(datetime2(2)
,dateadd(day, (row_number() over (order by (select 1)) -1)/2, @fromdate))
)
, [ThruDate]=dateadd(hour,19+12*((row_number() over (order by (select 1)) -1)%2)
,convert(datetime2(2)
,dateadd(day, (row_number() over (order by (select 1)) -1)/2, @fromdate))
)
from n as deka
cross join n as hecto /* 100 days */
--cross join n as kilo /* 2.73 years */
--cross join n as [tenK] /* 27.3 years */
order by 1
)
select
FromDate=convert(varchar(20),FromDate,120)
, ThruDate=convert(varchar(20),ThruDate,120)
, SumReading1=sum(Reading1)
, SumReading2=sum(Reading2)
from dates d
inner join t
on t.date >= d.fromdate
and t.date < d.thrudate
group by d.FromDate, d.ThruDate
order by d.FromDate, d.ThruDate
返回：
+---------------------+---------------------+-------------+-------------+
| FromDate | ThruDate | SumReading1 | SumReading2 |
+---------------------+---------------------+-------------+-------------+
| 2017-02-15 07:00:00 | 2017-02-15 19:00:00 | 392 | 364 |
| 2017-02-15 19:00:00 | 2017-02-16 07:00:00 | 101 | 93 |
| 2017-02-16 07:00:00 | 2017-02-16 19:00:00 | 34 | 31 |
+---------------------+---------------------+-------------+-------------+
答案 5 :(得分：0)
对于任何一天，我会用几个BETWEEN子查询来处理这个问题。类似的东西：
declare @refDate datetime
declare @midPeriod datetime
declare @endPeriod datetime
set @refDate = '2017-02-15 07:00'
set @midPeriod = dateadd(hh, 12, @refDate)
set @endPeriod = dateadd(hh, 24, @refDate)
select @refDate PeriodStart,
(select sum(Reading1) from #LocalTempTable where Date between @refDate and @midPeriod) EarlyReading1,
(select sum(Reading1) from #LocalTempTable where Date between @midPeriod and @endPeriod) LateReading1,
(select sum(Reading2) from #LocalTempTable where Date between @refDate and @midPeriod) EarlyReading2,
(select sum(Reading2) from #LocalTempTable where Date between @midPeriod and @endPeriod) LateReading2