Getting the 'date' out of a DateTime field in SQL Server - sql-server

Question

I have a date column where the date is displayed in the format 2009-11-18 10:55:28.370.
I just want to get the date (not time) out of that value. How do I do that?

Answer 1

If you're using SQL Server 2008, there is now a DATE datatype. Makes it a lot more natural!
SELECT CONVERT(Date, GETDATE())

Answer 2

it is called "flooring a datetime", do it like this to remove just the time (this is the fastest method, faster than using CONVERT() or CAST() sting formatting):
DECLARE #datetime datetime;
SET #datetime = '2008-09-17 12:56:53.430';
SELECT DATEADD(day,DATEDIFF(day,0,#datetime),0)
OUTPUT:
-----------------------
2008-09-17 00:00:00.000
(1 row(s) affected)
here is how to do it for other parts of a datetime:
--Floor a datetime
DECLARE #datetime datetime;
SET #datetime = '2008-09-17 12:56:53.430';
SELECT '0 None', #datetime -- none 2008-09-17 12:56:53.430
UNION SELECT '1 Second',DATEADD(second,DATEDIFF(second,'2000-01-01',#datetime),'2000-01-01') -- Second: 2008-09-17 12:56:53.000
UNION SELECT '2 Minute',DATEADD(minute,DATEDIFF(minute,0,#datetime),0) -- Minute: 2008-09-17 12:56:00.000
UNION SELECT '3 Hour', DATEADD(hour,DATEDIFF(hour,0,#datetime),0) -- Hour: 2008-09-17 12:00:00.000
UNION SELECT '4 Day', DATEADD(day,DATEDIFF(day,0,#datetime),0) -- Day: 2008-09-17 00:00:00.000
UNION SELECT '5 Month', DATEADD(month,DATEDIFF(month,0,#datetime),0) -- Month: 2008-09-01 00:00:00.000
UNION SELECT '6 Year', DATEADD(year,DATEDIFF(year,0,#datetime),0) -- Year: 2008-01-01 00:00:00.000
ORDER BY 1
PRINT' '
PRINT 'Note that when you are flooring by the second, you will often get an arithmetic overflow if you use 0. So pick a known value that is guaranteed to be lower than the datetime you are attempting to floor'
PRINT 'this always uses a date less than the given date, so there will be no arithmetic overflow'
SELECT '1 Second',DATEADD(second,DATEDIFF(second,DATEADD(day,DATEDIFF(day,0,#datetime),0)-1,#datetime),DATEADD(day,DATEDIFF(day,0,#datetime),0)-1) -- Second: 2008-09-17 12:56:53.000
OUTPUT:
-------- -----------------------
0 None 2008-09-17 12:56:53.430
1 Second 2008-09-17 12:56:53.000
2 Minute 2008-09-17 12:56:00.000
3 Hour 2008-09-17 12:00:00.000
4 Day 2008-09-17 00:00:00.000
5 Month 2008-09-01 00:00:00.000
6 Year 2008-01-01 00:00:00.000
(7 row(s) affected)
Note that when you are flooring by the second, you will often get an arithmetic overflow if you use 0. So pick a known value that is guaranteed to be lower than the datetime you are attempting to floor
this always uses a date less than the given date, so there will be no arithmetic overflow
-------- -----------------------
1 Second 2008-09-17 12:56:53.000
(1 row(s) affected)

Answer 3

If i got your question right,
select convert(varchar, creation_date , 103) as creation_date from tablename
Look at CAST and CONVERT

Answer 4

Here:
SELECT creation_date
FROM risks
WHERE creation_date = GETDATE()
This will return all creation_date values stored in the risks table that are exactly the same as what is returned by the GETDATE() function. I am assuming that the data type of creation_date is Date.

Answer 5

You just need to include creation_date in your select clause like this:
select id, creation_date from risks where creation_date = getdate()

Answer 6

You can always use the month/day/year functions to return it:
declare #date datetime
set #date = '1/1/10 12:00 PM'
select cast(month(#date) as varchar) + '/' + cast(day(#date) as varchar) + '/' + cast(year(#date) as varchar) as theDate