How to select date without time in SQL

Question

When I select date in SQL it is returned as 2011-02-25 21:17:33.933. But I need only the Date part, that is 2011-02-25. How can I do this?

Answer 1

For SQL Server 2008:

Convert(date, getdate())  

Please refer to https://learn.microsoft.com/en-us/sql/t-sql/functions/getdate-transact-sql

Answer 2

I guess he wants a string.

select convert(varchar(10), '2011-02-25 21:17:33.933', 120)

120 here tells the convert function that we pass the input date in the following format: yyyy-mm-dd hh:mi:ss.

Answer 3

Using CAST(GETDATE() As Date) worked for me

Answer 4

The fastest is datediff, e.g.

select dateadd(d, datediff(d,0, [datecolumn]), 0), other..
from tbl

But if you only need to use the value, then you can skip the dateadd, e.g.

select ...
WHERE somedate <= datediff(d, 0, getdate())

where the expression datediff(d, 0, getdate()) is sufficient to return today's date without time portion.

Answer 5

CAST(
        FLOOR( 
             CAST( GETDATE() AS FLOAT ) 
        )

AS DATETIME
)

http://www.bennadel.com/blog/122-Getting-Only-the-Date-Part-of-a-Date-Time-Stamp-in-SQL-Server.htm

Answer 6

For 2008 older version :

SELECT DATEADD(DAY, DATEDIFF(DAY, 0, GETDATE()), 0)

Answer 7

you can use like this

SELECT Convert(varchar(10), GETDATE(),120) 

Answer 8

In case if you need the time to be zeros like 2018-01-17 00:00:00.000:

SELECT CONVERT(DATETIME, CONVERT(DATE, GETDATE()), 121)

Answer 9

I would use DATEFROMPARTS function. It is quite easy and you don't need casting. As an example this query :

Select  DATEFROMPARTS(YEAR(GETDATE()), MONTH(GETDATE()), DAY(GETDATE())) as myNewDate

will return

2021-01-21

The good part you can also create you own date, for example you want first day of a month as a date, than you can just use like below:

Select  DATEFROMPARTS(YEAR(GETDATE()), MONTH(GETDATE()), 1) as myNewDate

The result will be:

2021-01-01

Answer 10

Its too late but following worked for me well

declare @vCurrentDate date=getutcdate()

select @vCurrentDate

When data type is date, hours would be truncated

Answer 11

You can try this one too.

SELECT CONVERT(DATE, GETDATE(), 120)

Answer 12

It's a bit late, but use the ODBC "curdate" function (angle brackes 'fn' is the ODBC function escape sequence).

SELECT {fn curdate()} 

Output: 2013-02-01

Answer 13

Convert it back to datetime after converting to date in order to keep same datatime if needed

select Convert(datetime, Convert(date, getdate())  )

Answer 14

select DATE(field) from table;

field value: 2020-12-15 12:19:00

select value: 2020-12-15

Answer 15

Use is simple:

convert(date, Btch_Time)

Example below:

Table:

Efft_d       Loan_I  Loan_Purp_Type_C   Orig_LTV    Curr_LTV    Schd_LTV    Un_drwn_Bal_a      Btch_Time            Strm_I  Btch_Ins_I
2014-05-31  200312500   HL03             NULL         1.0000    1.0000         1.0000      2014-06-17 11:10:57.330  1005    24851e0a-53983699-14b4-69109

Select * from helios.dbo.CBA_SRD_Loan where Loan_I in ('200312500') and convert(date, Btch_Time) = '2014-06-17'

Answer 16

If you want to return a date type as just a date use

CONVERT(date, SYSDATETIME())

or

SELECT CONVERT(date,SYSDATETIME()) 

or

DECLARE @DateOnly Datetime
SET @DateOnly=CONVERT(date,SYSDATETIME())

Answer 17

In PLSQL you can use

to_char(SYSDATE,'dd/mm/yyyy')

Answer 18

First Convert the date to float (which displays the numeric), then ROUND the numeric to 0 decimal points, then convert that to datetime.

convert(datetime,round(convert(float,orderdate,101),0) ,101)

Answer 19

Try this.

SELECT DATEADD(DD, 0, DATEDIFF(DD, 0, GETDATE()))

Answer 20

I would create a scalar function and use format () to set the datatype you want to see. It is must easy on the maintenance later.

Answer 21

Personal favorite:

select convert(datetime, convert(int, getdate()))