[英]SQL Server SUM(Values)
我有以下查询,效果很好。 查询将给定日期的值相加。
SELECT
SUM(fldValue) AS 'kWh',
DAY(fldDateTime) AS 'Day',
MONTH(fldDateTime) AS 'Month',
YEAR(fldDateTime) AS 'Year'
FROM
[Data.tblData]
WHERE
tblData_Id IN (SELECT DISTINCT tblData_Id
FROM [Data.tblData])
GROUP BY
YEAR(fldDateTime), MONTH(fldDateTime), DAY(fldDateTime),
tblData_Id,fldDateTime
ORDER BY
YEAR(fldDateTime), MONTH(fldDateTime), DAY(fldDateTime)
我的问题是它从午夜到午夜求和,我需要它在午夜(> =午夜)之后到第二天午夜之前求和。 这样做的原因是一天中输入的数据总是在午夜之后。 例如,第一个记录的数据将为“ 2016-01-01 00:01:00”,最后一个记录的数据将为“ 2016-01-02 00:00:00”。 这就是向我发送数据的硬件的工作方式。
我想知道如何在查询中封装> =午夜到午夜。
资料集:
DateTime Value
20/03/2016 00:30 69.00
20/03/2016 01:00 69.00
20/03/2016 01:30 69.00
20/03/2016 02:00 69.00
20/03/2016 02:30 69.00
20/03/2016 03:00 69.00
20/03/2016 03:30 11.88
20/03/2016 04:00 0.52
20/03/2016 04:30 1.51
20/03/2016 05:00 2.22
20/03/2016 05:30 2.11
20/03/2016 06:00 0.05
20/03/2016 06:30 6.78
20/03/2016 07:00 14.79
20/03/2016 07:30 1.57
20/03/2016 08:00 1.51
20/03/2016 08:30 4.81
20/03/2016 09:00 0.11
20/03/2016 09:30 8.99
20/03/2016 10:00 10.06
20/03/2016 10:30 15.28
20/03/2016 11:00 3.22
20/03/2016 11:30 1.73
20/03/2016 12:00 19.10
20/03/2016 12:30 2.08
20/03/2016 13:00 2.61
20/03/2016 13:30 0.84
20/03/2016 14:00 8.65
20/03/2016 14:30 2.37
20/03/2016 15:00 16.34
20/03/2016 15:30 12.66
20/03/2016 16:00 2.64
20/03/2016 16:30 0.19
20/03/2016 17:00 3.91
20/03/2016 17:30 2.39
20/03/2016 18:00 0.57
20/03/2016 18:30 1.30
20/03/2016 19:00 5.06
20/03/2016 19:30 17.45
20/03/2016 20:00 13.04
20/03/2016 20:30 5.00
20/03/2016 21:00 7.47
20/03/2016 21:30 5.09
20/03/2016 22:00 0.33
20/03/2016 22:30 5.29
20/03/2016 23:00 15.33
20/03/2016 23:30 5.39
21/03/2016 00:00 6.74
先感谢您。
2016年3月20日的预期总和输出值为:662.98
输出表如下所示:
SumValue Day Month Year Meter Id
659.18 20 3 2016 6
251.37 21 3 2016 6
279.03 22 3 2016 6
280.03 23 3 2016 6
284.22 24 3 2016 6
310.12 25 3 2016 6
320.84 26 3 2016 6
269.29 27 3 2016 6
276.11 28 3 2016 6
279.11 29 3 2016 6
值列是当天的值总和,由很多次独立时间组成。
首先,我不知道WHERE
子句在做什么,因此我将其删除。
其次,不要对列名使用单引号。
第三,您的GROUP BY
子句太复杂了。 您只需要在SELECT
包括未聚合的列。
最后,关键思想是在所有使用值的地方减去一小时。 这是一个简单的方法:
SELECT SUM(fldValue) AS kWh,
DAY(newdt) AS [Day],
MONTH(newdt) AS [Month],
YEAR(newdt) AS [Year]
FROM (SELECT d.*, DATEADD(hour, -1, fldDateTime) as newdt
FROM Data.tblData d
) d
GROUP BY YEAR(newdt), MONTH(newdt), DAY(newdt)
ORDER BY YEAR(newdt), MONTH(newdt), DAY(newdt)
答案与@Gordon相同,但您可以减去一分钟而不是一小时。
SELECT SUM(fldValue) AS kWh,
DAY(newdt) AS [Day],
MONTH(newdt) AS [Month],
YEAR(newdt) AS [Year]
FROM (SELECT d.*, DATEADD(minute, -1, fldDateTime) as newdt
FROM Data.tblData d
) d
GROUP BY YEAR(newdt), MONTH(newdt), DAY(newdt)
ORDER BY YEAR(newdt), MONTH(newdt), DAY(newdt)
declare @tempTable table ([DateTime] datetime, Value Float)
insert into @tempTable ([DateTime], [Value])
select convert(datetime,'20/03/2016 00:30',103), 69.00 union all
select convert(datetime,'20/03/2016 01:00',103), 69.00 union all
select convert(datetime,'21/03/2016 00:00',103), 6.74
select * from @tempTable
select [sum] = SUM(value), [year] = year(DT), [month] = month(DT), [day] = day(DT)
from (select Value, DT = dateadd(second, -1, [DateTime]) from @tempTable) x
group by year(DT), month(DT), day(DT)
使用以下查询将前一天的午夜值相加。
SELECT
SUM(fldValue) AS 'kWh',
CASE WHEN CONVERT(VARCHAR(8), fldDateTime, 108)='00:00:00' THEN DAY(fldDateTime)-1 ELSE DAY(fldDateTime) END AS 'Day',
MONTH(fldDateTime) AS 'Month',
YEAR(fldDateTime) AS 'Year'
FROM
Data.[tblData]
GROUP BY
YEAR(fldDateTime), MONTH(fldDateTime),CASE WHEN CONVERT(VARCHAR(8), fldDateTime, 108)='00:00:00' THEN DAY(fldDateTime)-1 ELSE DAY(fldDateTime) END
ORDER BY
YEAR(fldDateTime), MONTH(fldDateTime), CASE WHEN CONVERT(VARCHAR(8), fldDateTime, 108)='00:00:00' THEN DAY(fldDateTime)-1 ELSE DAY(fldDateTime) END
样本输出:
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.