让我们切入追逐。我有一个看起来像这样的表（使用SQL Server 2014）：
样本： 的 http://sqlfiddle.com/#!6/75f4a/1/0
CREATE TABLE TAB (
DT datetime,
VALUE float
);
INSERT INTO TAB VALUES
('2015-05-01 06:00:00', 12),
('2015-05-01 06:20:00', 10),
('2015-05-01 06:40:00', 11),
('2015-05-01 07:00:00', 14),
('2015-05-01 07:20:00', 15),
('2015-05-01 07:40:00', 13),
('2015-05-01 08:00:00', 10),
('2015-05-01 08:20:00', 9),
('2015-05-01 08:40:00', 5),
('2015-05-02 06:00:00', 19),
('2015-05-02 06:20:00', 7),
('2015-05-02 06:40:00', 11),
('2015-05-02 07:00:00', 9),
('2015-05-02 07:20:00', 7),
('2015-05-02 07:40:00', 6),
('2015-05-02 08:00:00', 10),
('2015-05-02 08:20:00', 19),
('2015-05-02 08:40:00', 15),
('2015-05-03 06:00:00', 8),
('2015-05-03 06:20:00', 8),
('2015-05-03 06:40:00', 8),
('2015-05-03 07:00:00', 21),
('2015-05-03 07:20:00', 12),
('2015-05-03 07:40:00', 7),
('2015-05-03 08:00:00', 10),
('2015-05-03 08:20:00', 4),
('2015-05-03 08:40:00', 10)
我需要：
换句话说，我希望有一个看起来像这样的表：
DATE | SUM VAL | ON HOUR
--------------------------
2015-03-01 | 24 | 8:00
2015-03-02 | 22 | 7:00
2015-03-03 | 24 | 6:00
前两点很容易（查看sqlfiddle）。我有第三个问题。我不能只喜欢选择Datepart（HOUR，DT），因为它必须被聚合。我试图使用JOINS和WHERE子句，但没有成功（表中可能会多次出现一些值，这会引发错误）。
我对SQL很新，但我遇到了困难。需要你的帮助！ ：）
答案 0 :(得分：2)
DECLARE @TAB TABLE
(
DT DATETIME ,
VALUE FLOAT
);
INSERT INTO @TAB
VALUES ( '2015-05-01 06:00:00', 12 ),
( '2015-05-01 06:20:00', 10 ),
( '2015-05-01 06:40:00', 11 ),
( '2015-05-01 07:00:00', 14 ),
( '2015-05-01 07:20:00', 15 ),
( '2015-05-01 07:40:00', 13 ),
( '2015-05-01 08:00:00', 10 ),
( '2015-05-01 08:20:00', 9 ),
( '2015-05-01 08:40:00', 5 ),
( '2015-05-02 06:00:00', 19 ),
( '2015-05-02 06:20:00', 7 ),
( '2015-05-02 06:40:00', 11 ),
( '2015-05-02 07:00:00', 9 ),
( '2015-05-02 07:20:00', 7 ),
( '2015-05-02 07:40:00', 6 ),
( '2015-05-02 08:00:00', 10 ),
( '2015-05-02 08:20:00', 19 ),
( '2015-05-02 08:40:00', 15 ),
( '2015-05-03 06:00:00', 8 ),
( '2015-05-03 06:20:00', 8 ),
( '2015-05-03 06:40:00', 8 ),
( '2015-05-03 07:00:00', 21 ),
( '2015-05-03 07:20:00', 12 ),
( '2015-05-03 07:40:00', 7 ),
( '2015-05-03 08:00:00', 10 ),
( '2015-05-03 08:20:00', 4 ),
( '2015-05-03 08:40:00', 10 );
WITH cteh
AS ( SELECT DT ,
CAST(dt AS DATE) AS D ,
SUM(VALUE) OVER ( PARTITION BY CAST(dt AS DATE),
DATEPART(hh, DT) ) AS S
FROM @TAB
),
ctef
AS ( SELECT * ,
ROW_NUMBER() OVER ( PARTITION BY D ORDER BY S ) AS rn
FROM cteh
)
SELECT D ,
S ,
CAST(DT AS TIME) AS H
FROM ctef
WHERE rn = 1
输出：
D S H
2015-05-01 24 08:00:00.0000000
2015-05-02 22 07:00:00.0000000
2015-05-03 24 06:00:00.0000000
答案 1 :(得分：1)
一种方法是使用具有最小每小时值的集合作为派生表并加入其中。我会做这样的事情：
;WITH CTE AS (
SELECT Cast(Format(DT, 'yyyy-MM-dd HH:00') AS datetime) AS DT, SUM(VALUE) AS VAL
FROM TAB
GROUP BY Format(DT, 'yyyy-MM-dd HH:00')
)
SELECT b.dt "Date", val "sum val", cast(min(a.dt) as time) "on hour"
FROM cte a JOIN (
SELECT Format(DT,'yyyy-MM-dd') AS DT, MIN(VAL) AS DAILY_MIN
FROM cte HOURLY
GROUP BY Format(DT,'yyyy-MM-dd')
) b ON CAST(a.DT AS DATE) = b.DT and a.VAL = b.DAILY_MIN
GROUP BY b.DT, a.VAL
这将得到：
Date sum val on hour
2015-05-01 24 08:00:00.0000000
2015-05-02 22 07:00:00.0000000
2015-05-03 24 06:00:00.0000000
我使用min（）作为时间部分，因为您的样本数据具有相同的低值，对于第3个单独的两个小时。如果你想要两个，那么从外部select和group by中删除min函数。然后你会得到：
Date sum val on hour
2015-05-01 24 08:00:00.0000000
2015-05-02 22 07:00:00.0000000
2015-05-03 24 06:00:00.0000000
2015-05-03 24 08:00:00.0000000
我确信它可以改进，但你应该明白这一点。
答案 2 :(得分：1)
这是一种方法，它使用临时表（与其他解决方案中的CTE相对）来存储计算值，然后过滤结果以获得所需的输出：
-- INSERT CALCULATED GROUPED VALUES INTO TEMP TABLE
SELECT CONVERT(DATE, DT) AS DateVal ,
SUM(VALUE) AS SumVal ,
DATEPART(HOUR, CONVERT(TIME, DT)) AS HourVal
INTO #TEMP_CALC
FROM TAB
GROUP BY CONVERT(DATE, DT) , DATEPART(HOUR, CONVERT(TIME, DT))
-- TAKE THE RELEVANT ROWS
SELECT t.DateVal ,
MIN(t.SumVal) AS SumVal ,
( SELECT TOP 1
HourVal
FROM #TEMP_CALC t2
WHERE t2.DateVal = t.DateVal
AND t2.SumVal = MIN(t.SumVal)
) AS MinHour
FROM #TEMP_CALC t
GROUP BY t.DateVal
ORDER BY DateVal
答案 3 :(得分：0)
您可以使用DATEDIFF以小时和天为单位从任何起始时间点（此示例中为1990-1-1）获取时间跨度。跨越用于分组和排序的用法，最后使用具有相同起点的DATEADD来重建它：
WITH dates AS (
SELECT CAST(DT AS DATETIME) AS Date, -- cast the value to date
value FROM dbo.TAB AS T
),
ddh AS (SELECT
date,
DATEDIFF(DAY, '1990-1-1', date) AS daySpan, -- days span
DATEDIFF(HOUR, '1990-1-1', date) AS hourSpan, -- hours span
value
FROM dates
),
ddhv AS ( SELECT
daySpan,
hourSpan,
SUM(value) AS sumValues -- sum...
FROM ddh
group BY daySpan, hourSpan -- ...grouped by day & hour
),
ddhvr AS ( SELECT
daySpan,
hourSpan,
sumValues,
-- number rows by hourly sum of the value
ROW_NUMBER() OVER (PARTITION BY daySpan ORDER BY sumValues) AS row
FROM ddhv
)
SELECT
DATEADD(HOUR, hourSpan, '1990-1-1') AS DayHour, -- rebuild the date/hour
sumValues
FROM ddhvr
WHERE row = 1 -- take only the first occurrence for each day
此查询的优点是您可以轻松更改句点和起点。例如，您可以在6:30 AM而不是00:00开始您的日期，以便比较时间段为6:30至7:30，7：30至8:30并继续。您也可以更改分组单位，例如，而不是1小时，它可能是半小时，或5分钟或2小时。如果您需要这样做，请see this SO answer。在那里，您将看到如何在不同时期进行分组，并回到期间的起始点。这只是一些简单的数学。
答案 4 :(得分：0)
我测试了我的小提琴：
with agg as (
select cast(dt as date) as dt, datepart(hh, dt) as hr, sum(VALUE) as sum_val
from TAB
group by cast(dt as date), datepart(hh, dt)
)
select
dt, min(sum_val) as "SUM VAL",
(
select cast(hr as varchar(2)) + ':00' from agg as agg2
where agg2.dt = agg.dt and not exists (
/* select earliest in case of ties */
select 1 from agg as agg3
where agg3.dt = agg2.dt and agg3.sum_val >= agg3.sum_val and agg3.hr > agg2.hr
)
) as "ON HOUR"
from agg
group by dt;