June 16, 2009 at 6:31 am
hi,
i have two date in sql server 2005
1/12/2008 + hours
31/01/2009 + hours
i'd like in a query sql to know the number of day in year 2009 and the number of day total !
Someone can help me, i'm a beginner !
Have a nice day
Christophe
June 16, 2009 at 7:20 am
You can use datediff to get the number of days;
select datediff(dd,'2001-01-01 00:00:00','2001-12-31 00:00:00')
June 26, 2009 at 12:14 am
Could you explain your problem in a little more detail and perhaps show us what you are looking for with a couple of examples?
Your initial post is a little vaque as to what you are looking for.
June 26, 2009 at 1:45 am
Hi all,
i'm a beginner in t sql and i have a table with some field, text and date
i have in one row two date
dateStart and dateEnd
22/10/2008 19/01/2009
and IN ONE query i'd like to have :
- the total of day BETWEEN the two date
- the total day for dateEnd at begin 1/01/2009 because we are in 2009
Thanks for your time
Christ
June 26, 2009 at 2:14 am
I'm assuming the second requirement is for the day number of the EndDate relative to the current year as in a Julian date. Therefore try:
SELECTDATEDIFF(DD,dateStart,DateEnd) AS DaysDiff,
DATEDIFF(DD,CONVERT(datetime,CONVERT(char(4),YEAR(dateEnd)) + '0101'),dateEnd) + 1AS DayOfYear
FROM.........
Mike
June 26, 2009 at 8:47 am
Is this what you are looking for?
--dateStart and dateEnd
--22/10/2008 19/01/2009
--
--and IN ONE query i'd like to have :
--
--- the total of day BETWEEN the two date
--- the total day for dateEnd at begin 1/01/2009 because we are in 2009
create table #TestTab (
RecID int identity(1,1),
DateStart datetime,
DateEnd datetime);
insert into #TestTab (DateStart, DateEnd)
select '2008-10-22', '2009-01-19' union all
select '2008-12-31', '2009-02-01' union all
select '2009-01-01', '2009-01-01';
select
DateStart,
DateEnd,
datediff(dd, DateStart, DateEnd) + 1 as TotalDays, -- assumes 1 if dates are the same
datediff(dd, dateadd(yy, datediff(yy, 0, DateEnd), 0), DateEnd) + 1 as DaysInCurrentYear -- same as above
from
#TestTab;
drop table #TestTab;
July 2, 2009 at 3:48 am
Hi all,
thanks for your time, that's excatly what i want !
thanks for all
Christophe
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