Get day of week in SQL Server 2005/2008 - sql

Question

If I have a date 01/01/2009, I want to find out what day it was e.g. Monday, Tuesday, etc...
Is there a built-in function for this in SQL Server 2005/2008? Or do I need to use an auxiliary table?

Answer 1

Use DATENAME or DATEPART:
SELECT DATENAME(dw,GETDATE()) -- Friday
SELECT DATEPART(dw,GETDATE()) -- 6

Answer 2

Even though SQLMenace's answer has been accepted, there is one important SET option you should be aware of
SET DATEFIRST
DATENAME will return correct date name but not the same DATEPART value if the first day of week has been changed as illustrated below.
declare #DefaultDateFirst int
set #DefaultDateFirst = ##datefirst
--; 7 First day of week is "Sunday" by default
select [#DefaultDateFirst] = #DefaultDateFirst
set datefirst #DefaultDateFirst
select datename(dw,getdate()) -- Saturday
select datepart(dw,getdate()) -- 7
--; Set the first day of week to * TUESDAY *
--; (some people start their week on Tuesdays...)
set datefirst 2
select datename(dw,getdate()) -- Saturday
--; Returns 5 because Saturday is the 5th day since Tuesday.
--; Tue 1, Wed 2, Th 3, Fri 4, Sat 5
select datepart(dw,getdate()) -- 5 <-- It's not 7!
set datefirst #DefaultDateFirst

Answer 3

SELECT CASE DATEPART(WEEKDAY,GETDATE())
WHEN 1 THEN 'SUNDAY'
WHEN 2 THEN 'MONDAY'
WHEN 3 THEN 'TUESDAY'
WHEN 4 THEN 'WEDNESDAY'
WHEN 5 THEN 'THURSDAY'
WHEN 6 THEN 'FRIDAY'
WHEN 7 THEN 'SATURDAY'
END

Answer 4

EUROPE:
declare #d datetime;
set #d=getdate();
set #dow=((datepart(dw,#d) + ##DATEFIRST-2) % 7+1);

Answer 5

To get a deterministic value for the day of week for a given date you could use a combination of DATEPART() and ##datefirst. Otherwise your dependent on the settings on the server.
Check out the following site for a better solution:
MS SQL: Day of Week
The day of week will then be in the range 0 to 6, where 0 is Sunday, 1 is Monday, etc. Then you can use a simple case statement to return the correct weekday name.

Answer 6

With SQL Server 2012 and onward you can use the FORMAT function
SELECT FORMAT(GETDATE(), 'dddd')

Answer 7

You can use DATEPART(dw, GETDATE()) but be aware that the result will rely on SQL server setting ##DATEFIRST value which is the first day of week setting (In Europe default value 7 which is Sunday).
If you want to change the first day of week to another value, you could use SET DATEFIRST but this may affect everywhere in your query session which you do not want.
Alternative way is to explicitly specify the first day of week value as parameter and avoid depending on ##DATEFIRST setting. You can use the following formula to achieve that when need it:
(DATEPART(dw, GETDATE()) + ##DATEFIRST + 6 - #WeekStartDay) % 7 + 1
where #WeekStartDay is the first day of the week you want for your system (from 1 to 7 which means from Monday to Sunday).
I have wrapped it into below function so we can reuse it easily:
CREATE FUNCTION [dbo].[GetDayInWeek](#InputDateTime DATETIME, #WeekStartDay INT)
RETURNS INT
AS
BEGIN
--Note: #WeekStartDay is number from [1 - 7] which is from Monday to Sunday
RETURN (DATEPART(dw, #InputDateTime) + ##DATEFIRST + 6 - #WeekStartDay) % 7 + 1
END
Example usage:
GetDayInWeek('2019-02-04 00:00:00', 1)
It is equivalent to following (but independent to SQL server DATEFIRST setting):
SET DATEFIRST 1
DATEPART(dw, '2019-02-04 00:00:00')

Answer 8

this is a working copy of my code check it, how to retrive day name from date in sql
CREATE Procedure [dbo].[proc_GetProjectDeploymentTimeSheetData]
#FromDate date,
#ToDate date
As
Begin
select p.ProjectName + ' ( ' + st.Time +' '+'-'+' '+et.Time +' )' as ProjectDeatils,
datename(dw,pts.StartDate) as 'Day'
from
ProjectTimeSheet pts
join Projects p on pts.ProjectID=p.ID
join Timing st on pts.StartTimingId=st.Id
join Timing et on pts.EndTimingId=et.Id
where pts.StartDate >= #FromDate
and pts.StartDate <= #ToDate
END

Answer 9

If you don't want to depend on ##DATEFIRST or use DATEPART(weekday, DateColumn), just calculate the day of the week yourself.
For Monday based weeks (Europe) simplest is:
SELECT DATEDIFF(day, '17530101', DateColumn) % 7 + 1 AS MondayBasedDay
For Sunday based weeks (America) use:
SELECT DATEDIFF(day, '17530107', DateColumn) % 7 + 1 AS SundayBasedDay
This return the weekday number (1 to 7) ever since January 1st respectively 7th, 1753.

Answer 10

In addition all with above answers, dw stands for Week Day
SELECT DATENAME(WEEKDAY,GETDATE()) AS WeekDay
or
SELECT DATENAME(W,GETDATE()) AS WeekDay

Answer 11

You may find this version useful for returning a shortened weekday name in one line.
-- Test DATA
select ##datefirst
create table #test (datum datetime)
insert #test values ('2013-01-01')
insert #test values ('2013-01-02')
insert #test values ('2013-01-03')
insert #test values ('2013-01-04')
insert #test values ('2013-01-05')
insert #test values ('2013-01-06')
insert #test values ('2013-01-07')
insert #test values ('2013-01-08')
-- Test DATA
select Substring('Sun,Mon,Tue,Wed,Thu,Fri,Sat,Sun,Mon,Tue,Wed,Thu,Fri,Sat',
(DATEPART(WEEKDAY,datum)+##datefirst-1)*4+1,3),Datum
from #test