Say, I have two DATETIME2 columns and I need to select dates between the two. Example: 1/1/2012 08:00 and 1/1/2012 09:00 should produce 1/1/2012 08:30.
说,我有两个DATETIME2列,我需要选择两者之间的日期。示例:1/1/2012 08:00和1/1/2012 09:00应该生成1/1/2012 08:30。
I'm trying this:
我正在尝试这个:
SELECT CAST((CAST(dtOut AS float(53)) +
CAST(dtIn AS float(53))) / 2 AS DATETIME2) FROM t;
But I get an error that explicit conversion from DATETIME2 is not allowed.
但是我得到一个错误,即不允许从DATETIME2进行显式转换。
Any idea how to do it?
知道怎么做吗?
4 个解决方案
#1
3
DateDiff will find the difference between two dates.
DateDiff将找到两个日期之间的差异。
select Dateadd(n, DATEDIFF(n, dtIn, dtOut)/2, dtIn) FROM t
As an aside, your method would work if the fields were datetime, not datetime2.
另外,如果字段是datetime而不是datetime2,那么您的方法将起作用。
#2
0
Try Out this.
试试这个。
select Dateadd(n, DATEDIFF(n, '1/1/2012 08:00', '1/1/2012 09:00')/2, '1/1/2012 08:00');
#3
0
declare @d1 datetime2, @d2 datetime2
select @d1='1/1/2012 08:00',@d2='1/1/2012 09:00'
select dateadd(minute,datediff(minute ,@d1,@d2)/2.0,@d1)
#4
0
select dateadd(SECOND,datediff(SECOND ,dtIn,dtOut)/2.0,dtIn)
选择dateadd(SECOND,datediff(SECOND,dtIn,dtOut)/2.0,dtIn)
#1
3
DateDiff will find the difference between two dates.
DateDiff将找到两个日期之间的差异。
select Dateadd(n, DATEDIFF(n, dtIn, dtOut)/2, dtIn) FROM t
As an aside, your method would work if the fields were datetime, not datetime2.
另外,如果字段是datetime而不是datetime2,那么您的方法将起作用。
#2
0
Try Out this.
试试这个。
select Dateadd(n, DATEDIFF(n, '1/1/2012 08:00', '1/1/2012 09:00')/2, '1/1/2012 08:00');
#3
0
declare @d1 datetime2, @d2 datetime2
select @d1='1/1/2012 08:00',@d2='1/1/2012 09:00'
select dateadd(minute,datediff(minute ,@d1,@d2)/2.0,@d1)
#4
0
select dateadd(SECOND,datediff(SECOND ,dtIn,dtOut)/2.0,dtIn)
选择dateadd(SECOND,datediff(SECOND,dtIn,dtOut)/2.0,dtIn)