我有 SQL 表,例如 DateDiff 的格式为 (hh.mm)
DateDiff ATM
1.45 121
1.50 121
1.50 121
当我在 ATM 上按 Group by
对 DateDiff 进行 Sum
时,它会显示如下结果
4.45 121
但实际的日期时间差应该是
5.25 121
我们如何在 SQL 查询中实现相同的效果
Select Sum(Cast(Convert(Varchar(10),DATEDIFF(HOUR,TicketRaisedOn,ClosedOn))+'.'+
Convert(Varchar(10),DATEDIFF(MINUTE,TicketRaisedOn,ClosedOn)%60) as Numeric(18,2)))[Down Time],ATM
From Ticket Where Closed=1 And DATEPART(DAYOFYEAR,GETDATE())=DATEPART(DAYOFYEAR,TicketRaisedOn)
Group BY ATM Order By [Down Time] Desc
TicketRaishedOn
和 ClosedOn
为 DateTime
数据库是SQL Server 2008
上面的查询将打印这样的结果(但它是错误的,因为它会将其求和为数字而不是日期时间格式)
Down Time ATM
16.95 282
14.46 1811
14.20 52
14.04 936
示例数据
Select TicketRaisedOn,ClosedOn,ATM
From Ticket
Where ATM=282 And DATEPART(DAYOFYEAR,GETDATE())=DATEPART(DAYOFYEAR,TicketRaisedOn)
And Closed=1
TicketRaisedOn ClosedOn ATM
2012-12-21 01:15:23.793 2012-12-21 15:11:59.240 282
2012-12-21 16:42:29.820 2012-12-21 18:21:30.797 282
最佳答案
在格式化之前进行求和
SELECT
ATM,
CONVERT(VARCHAR(10), SUM(DATEDIFF(Minute, TicketRaisedOn, ClosedOn)) / 60)
+ '.' +
RIGHT('00' + CONVERT(VARCHAR(2), SUM(DATEDIFF(Minute, TicketRaisedOn, ClosedOn)) % 60), 2)
FROM Ticket
GROUP BY ATM
Sql fiddle :http://sqlfiddle.com/#!3/eca01/1
关于sql - sql中日期时间差的总和(HH.MM),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/13991440/