Question

我试图将大量数据分组到较小的包中。

目前我的查询代码如下

SELECT [DateTime] ,[KW] FROM [POWER] WHERE datetime >= '2014-04-14 06:00:00' and datetime < '2014-04-21 06:00:00' ORDER BY datetime

给了我

DateTime KW 4/14/2014 6:00:02.0 1947 4/14/2014 6:00:15.0 1946 4/14/2014 6:00:23.0 1947 4/14/2014 6:00:32.0 1011 4/14/2014 6:00:43.0 601 4/14/2014 6:00:52.0 585 4/14/2014 6:01:02.0 582 4/14/2014 6:01:12.0 580 4/14/2014 6:01:21.0 579 4/14/2014 6:01:32.0 579 4/14/2014 6:01:44.0 578 4/14/2014 6:01:53.0 578 4/14/2014 6:02:01.0 577 4/14/2014 6:02:12.0 577 4/14/2014 6:02:22.0 577 4/14/2014 6:02:32.0 576 4/14/2014 6:02:42.0 578 4/14/2014 6:02:52.0 577 4/14/2014 6:03:02.0 577 4/14/2014 6:03:12.0 577 4/14/2014 6:03:22.0 578 . . . . 4/21/2014 5:59:55.0 11

现在变电站每隔10秒就有一个读数。现在我想将这些数据分组为每小时读数。

因此00:00-01:00 = sum([KW]] for where datetime >= '^date^ 00:00:00' and datetime < '^date^ 01:00:00'

我尝试使用转换将日期时间更改为日期和时间字段，然后仅将所有时间字段添加到一起，但没有成功。

有人可以帮助我，我不确定这样做的正确方法。感谢

已添加

好的，所以Datetime之间的溢出效果很好，但好像我添加了一个SUM（[KW]）函数，SQL给出了一个错误。如果我包含任何组功能，它也会唠叨。

下面是有效的，我仍然需要按小时组合总和。

我已尝试使用Group By Hour和Group by DATEPART(Hour,[DateTime])

两者都没有用。

SELECT DATEPART(Hour,[DateTime]) Hour ,DATEPART(Day,[DateTime]) Day ,DATEPART(Month,[DateTime]) Month ,([KVAReal]) ,([KVAr]) ,([KW]) FROM [POWER].[dbo].[IT10t_PAC3200] WHERE datetime >= '2014-04-14 06:00:00' and datetime < '2014-04-21 06:00:00' order by datetime

Answer 1

功能convert(varchar(13), getdate(), 120)显示2014-06-03 16。您可以使用它按小时分组：

SELECT  convert(varchar(13), [DateTime], 120) as dt
,       SUM(KW) as SumKwPerHour
FROM    POWER
WHERE   [DateTime] >= '2014-04-14 06:00:00' 
        AND [DateTime]< '2014-04-21 06:00:00'
GROUP BY
        convert(varchar(13), [DateTime], 120)
ORDER BY
        dt

Answer 2

好的，这是适用于我的解决方案。

Declare @Begin Varchar(60), @End Varchar(60) Set @Begin = '2014-05-22 06:00:00' Set @End = '2014-06-01 06:00:00'

SELECT ID='10T' ,DATEPART(month,[DateTime]) Month ,DATEPART(day,[DateTime]) Day ,DATEPART(hour,[DateTime]) as Hour ,avg([kw]) hourly_kWh_10T ,avg([KVAr]) hourly_kVarh_10T ,avg([KVAReal]) hourly_kVAh_10T ,(case when(DATEPART(hour,[DateTime]) <=6 and DATEPART(hour,[DateTime]) >18) then 'D' else 'N' end) shift FROM [POWER] where DateTime <= @Begin and DateTime > @End group by DATEPART(Hour,[DateTime]),DATEPART(Day,[DateTime]),DATEPART(Month,[DateTime])

这段代码给了我这个我想要的结果。我还包括一个变量起点来减少不同日期的输入。 +添加了一个if函数（Case when），以确定在白天或夜班期间是否消耗了电量。

ID Month Day Hour hourly_kWh_10T hourly_kVarh_10T hourly_kVAh_10T shift 10T 5 22 6 269.278551 80.771587 294.038997 D 10T 5 22 7 241.213296 75.991689 268.085872 D 10T 5 22 8 283.925 93.302777 319.211111 D 10T 5 22 9 11.763888 31.313888 36.372222 D 10T 5 22 10 215.947222 69.702777 243.541666 D 10T 5 22 11 1895.816666 396.805555 1948.061111 D 10T 5 22 12 2385.486033 513.589385 2447.648044 D 10T 5 22 13 440.737569 126.209944 475.049723 D 10T 5 22 14 737.158333 183.05 775.763888 D 10T 5 22 15 41.961111 38.086111 67.277777 D 10T 5 22 16 11.875 30.577777 35.736111 D 10T 5 22 17 11.263888 27.563888 32.497222 D 10T 5 22 18 11.104956 26.381924 31.323615 N 10T 5 22 19 11.648936 28.813829 34.015957 N 10T 5 22 20 229.819944 75.227146 268.432132 N 10T 5 22 21 300.597222 92.661111 340.413888 N 10T 5 22 22 494.575 124.358333 527.183333 N 10T 5 22 23 922.244444 190.472222 954.961111 N 10T 5 23 0 2445.908333 516.008333 2507.613888 N 10T 5 23 1 1399.147222 317.380555 1446.786111 N 10T 5 23 2 258.097222 81.641666 288.308333 N 10T 5 23 3 258.480555 79.694444 285.488888 N 10T 5 23 4 262.108333 82.455555 290.261111 N 10T 5 23 5 270.830555 82.030555 297.011111 N 10T 5 23 6 570.836111 151.930555 606.05 D 10T 5 23 7 10.580555 24.488888 29.233333 D

在SQL查询中求和行的时间范围

2 个答案: