I converted some string values to datetime
as below:
'20131018'
'20130917'
by using
CONVERT(DATETIME, @date, 101)
I got the following
'2013-10-18 00:00:00.000'
'2013-09-17 00:00:00.000'
How would I remove the timepart of the values?
I converted some string values to datetime
as below:
'20131018'
'20130917'
by using
CONVERT(DATETIME, @date, 101)
I got the following
'2013-10-18 00:00:00.000'
'2013-09-17 00:00:00.000'
How would I remove the timepart of the values?
SUBSTRING ( expression ,start , length ) where expression will be your timestamp string and start will be 1 and length will be length of "yyyy-mm-dd" i.e. 10. for more reference please check http://technet.microsoft.com/en-us/library/ms187748.aspx
Check what said by "marc_s". Just use DATE instead of DATETIME.This will be better option.
Have some code I made. If you can figure out how to make this faster or better let me know! ;)
case --Format date into YYYYMMDD with placeholder 0's
when (CAST(DATEPART(DD,DATEADD(day, 7, <DATE>)) as int ) < 10) AND (CAST(DATEPART(MM,DATEADD(day, 7, <DATE>)) as int) < 10)
then CAST(DATEPART(YYYY,DATEADD(day, 7, <DATE>)) as CHAR(4))
+'0'+ LTRIM(RTRIM(CAST(DATEPART(MM,DATEADD(day, 7, <DATE>)) as CHAR(2))))
+'0'+LTRIM(RTRIM(CAST(DATEPART(DD,DATEADD(day, 7, <DATE>)) as CHAR(2))))
when (CAST(DATEPART(MM,DATEADD(day, 7, <DATE>)) as int) < 10)
then CAST(DATEPART(YYYY,DATEADD(day, 7, <DATE>)) as CHAR(4))
+'0'+LTRIM(RTRIM(CAST(DATEPART(MM,DATEADD(day, 7, <DATE>)) as CHAR(2))))
+CAST(DATEPART(DD,DATEADD(day, 7, <DATE>)) as CHAR(2))
when (CAST(DATEPART(DD, DATEADD(day, 7, <DATE>)) as int ) < 10)
then CAST(DATEPART(YYYY,DATEADD(day, 7, <DATE>)) as CHAR(4))
+ CAST(DATEPART(MM,DATEADD(day, 7, <DATE>)) as CHAR(2))
+'0'+LTRIM(RTRIM(CAST(DATEPART(DD,DATEADD(day, 7, <DATE>)) as CHAR(2))))
else CAST(DATEPART(YYYY,DATEADD(day, 7, <DATE>)) as CHAR(4))
+CAST(DATEPART(MM,DATEADD(day, 7, <DATE>)) as CHAR(2))
+CAST(DATEPART(DD,DATEADD(day, 7, <DATE>)) as CHAR(2))
end
Here's how it works. Put that after SELECT. Do a Find and Replace with and put your date in there. It checks the number of digits in the month and day, then inserts a 0 or two into the string when appropriate. I saved this block of code so that it would be easy to use when I encounter this problem. Hope it helps!
--EDIT-- It occurs to me, this might be the reverse of what you want to do...