Get day of week in SQL Server 2005/2008

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?

EUROPE:

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

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

With SQL Server 2012 and onward you can use the `FORMAT` function

SELECT FORMAT(GETDATE(), 'dddd')

Use [`DATENAME`](

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) or [`DATEPART`](

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):

SELECT DATENAME(dw,GETDATE()) -- Friday
SELECT DATEPART(dw,GETDATE()) -- 6

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.

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')

To get a deterministic value for the day of week for a given date you could use a combination of [DATEPART()][1] and [@@datefirst][2]. Otherwise your dependent on the settings on the server.

Check out the following site for a better solution:
[MS SQL: Day of Week][3]

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.


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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

Even though [SQLMenace's answer][1] has been accepted, there is one important `SET` option you should be aware of

> [SET DATEFIRST][2]

[DATENAME][3] will return correct date *name* but not the same [DATEPART][4] 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


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