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总计不同日期的总和

[英]Sum the different date total

我的sql表中有以下数据:

 Tran_Date          |  Amount 
2013-05-01 20:09:49 | 50.00
2013-05-02 04:09:49 | 50.00
2013-05-02 20:09:49 | 500.00

我想在第二天凌晨5点之前将金额总计。 结果应该如下。

Amount 
100.00 
500.00

我尝试了以下代码,但结果错误:

SELECT DATEADD(hh, 5, DATEADD(dd, DATEDIFF(dd, 0, DATEADD(dd,1,f.TRAN_DATE)) AS sDate, 
       SUM(Amount) 
FROM TRAN_TABLE 
GROUP BY sDate

如何做到这一点? 谢谢

对于任何SQL Server版本

select [Date]=DATEADD(DAY,DATEDIFF(DAY,0,DATEADD(HOUR,-5,Tran_Date)),0),
       Total=SUM(Amount)
from tbl
group by DATEADD(DAY,DATEDIFF(DAY,0,DATEADD(HOUR,-5,Tran_Date)),0)
order by [Date];

对于SQL Server 2008+,可以使用DATE数据类型

select [Date]=CAST(DATEADD(HOUR,-5,Tran_Date) as date),
       Total=SUM(Amount)
from tbl
group by CAST(DATEADD(HOUR,-5,Tran_Date) as date)
order by [Date];

您需要减去5小时后提取一个DATE,然后按以下方式分组:

SELECT
    CAST(DATEADD(hour, -5, TRAN_DATE) AS DATE) AS sDate,
    SUM(Amount)
FROM TRAN_TABLE
GROUP BY CAST(DATEADD(hour, -5, TRAN_DATE) AS DATE)

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