Question

我有一个问题，我的SQL查询我也写了。很久以前，因为我写了一个查询，所以我可以使用我的一些帮助。我试过寻找例子，但没有找到正确的结果。我写了一个查询，但它真的不适合我.. 我想要做的是获取数据库中每个日期的总功耗之和。我的表看起来像：

|HistoryProbes|
|-------------|
|id (int) pk  |
|timestamp (datetime)  formatted as: "yyyy-MM-ddTHH:mm:ss"|
|PowerConsumption (int)|

我找到了一个做得很好的样品..但它不是我最好的解决方案.. 它可以在http://cephas.net/blog/2005/12/06/sql-server-group-by-datetime/

找到

到目前为止，我有这个工作

SELECT distinct CONVERT(varchar, timestamp, 111) AS thedate
FROM         HistoryProbes

我得到了价值25/11/2009和24/11/2009，但我无法获得PowerConsumption的总和

感谢。

Answer 1

select CONVERT(varchar, timestamp, 111) as timestamp_by_day
, sum(PowerConsumption) as total_power
from HistoryProbes
group by CONVERT(varchar, timestamp, 111)
order by CONVERT(varchar, timestamp, 111)

Answer 2

这样的事情会给你每天的总和

DECLARE @HistoryProbes TABLE(
        id INT,
        timesmp  DATETIME,
        PowerConsumption INT
)

INSERT INTO @HistoryProbes (id,timesmp,PowerConsumption) SELECT 1, '01 Jan 2009 12:00:00',1
INSERT INTO @HistoryProbes (id,timesmp,PowerConsumption) SELECT 2, '01 Jan 2009 11:00:00',2
INSERT INTO @HistoryProbes (id,timesmp,PowerConsumption) SELECT 3, '01 Jan 2009 13:00:00',3
INSERT INTO @HistoryProbes (id,timesmp,PowerConsumption) SELECT 4, '01 Jan 2009 14:00:00',4
INSERT INTO @HistoryProbes (id,timesmp,PowerConsumption) SELECT 5, '02 Jan 2009 12:00:00',14
INSERT INTO @HistoryProbes (id,timesmp,PowerConsumption) SELECT 6, '02 Jan 2009 11:00:00',24
INSERT INTO @HistoryProbes (id,timesmp,PowerConsumption) SELECT 7, '03 Jan 2009 13:00:00',34
INSERT INTO @HistoryProbes (id,timesmp,PowerConsumption) SELECT 8, '03 Jan 2009 14:00:00',44

SELECT  DATEADD(dd,0, DATEDIFF(dd,0,timesmp)),
        SUM(PowerConsumption)
FROM    @HistoryProbes
GROUP BY DATEADD(dd,0, DATEDIFF(dd,0,timesmp))

Answer 3

试试这个：
SELECT Convert(varchar, timestamp, 111) as thedate, SUM(PowerConsumption) as Total FROM HistoryProbes GROUP BY Convert(varchar, timestamp, 111)

我不确定为什么你需要在那里分明;因为你没有加入任何其他表

日期选择查询按天分组

3 个答案: