sql中日期时间差异的总和（HH.MM）

Question

我有 SQL 表，例如 DateDiff 的格式（hh.mm）

DateDiff   ATM
1.45       121
1.50       121
1.50       121

当我在 ATMSum上使用 DateDiff 时，Group by它会显示如下结果

4.45    121

但实际的日期时差应该是

5.25     121

我们如何在 SQL Query 中实现相同的功能

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

TicketRaisedOn&ClosedOn属于日期时间

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，因为它会将其总结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

样本数据

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

Answer 1

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 小提琴：http://sqlfiddle.com/#!3/eca01 /1