The day of the week can be retrieved in SQL Server by using the DatePart function. The value returned by function is between 1 (Sunday) and 7 (Saturday). To convert this to a string representing the day of the week, use a CASE statement.
Method 1:
Create function running following script:
CREATE FUNCTION dbo.udf_DayOfWeek(@dtDate DATETIME)
RETURNS VARCHAR(10)
AS
BEGIN
DECLARE @rtDayofWeek VARCHAR(10)
SELECT @rtDayofWeek = CASE DATEPART(weekday,@dtDate)
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
RETURN (@rtDayofWeek)
END
GO
Call this function like this:
SELECT dbo.udf_DayOfWeek(GETDATE()) AS DayOfWeek
ResultSet:
DayOfWeek
———-
Monday
Method 2: (This is update from comments I received below)
SELECT DATENAME(dw, GETDATE())
Reference : Pinal Dave (http://blog.SQLAuthority.com)
why wouldn’t you use the DATENAME function?
Why not use the datename() function instead – it’s already built in to SQL Server?
IE – SELECT DATENAME(dw, GETDATE())
Chasin and Simon,
Thank you for brining up the DATENAME function.
I have modified my post to reflect that.
Regards,
Pinal Dave
CREATE FUNCTION dbo.udf_DayOfWeek(@dtDate DATETIME)
RETURNS VARCHAR(10)
AS
BEGIN
DECLARE @rtDayofWeek VARCHAR(10)
SELECT @rtDayofWeek = CASE DATEPART(weekday,@dtDate)
WHEN 0 THEN ‘Sunday’
WHEN 1 THEN ‘Monday’
WHEN 2 THEN ‘Tuesday’
WHEN 3 THEN ‘Wednesday’
WHEN 4 THEN ‘Thursday’
WHEN 5 THEN ‘Friday’
WHEN 6 THEN ‘Saturday’
END
RETURN (@rtDayofWeek)
END
GO
for right result we have to change this to
CREATE FUNCTION dbo.udf_DayOfWeek(@dtDate DATETIME)
RETURNS VARCHAR(10)
AS
BEGIN
DECLARE @rtDayofWeek VARCHAR(10)
SELECT @rtDayofWeek = CASE DATEPART(weekday,@dtDate)
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
RETURN (@rtDayofWeek)
END
GO
Thanks Mahesh,
I guess I had offset of 1 digit. I have reflected the changes to above post.
Regards,
Pinal Dave (http://www.SQLAuthority.com)
Why would you even keep the first method up on the site when there is a built in function that performs the same job? It doesn’t seem like a best practice to rewrite working functions on your SQL Server.
Steve,
This is fun article and never claimed as best practice. I kept the first method up cause I think it is still interesting to read it.
Regards,
Pinal Dave (SQLAuthority.com)
The tricky thing here is that Sunday does not always equal 1. The values returned by DATEPART(dw) will change based on the setting of @@DATEFIRST (which indicates the first day of the week). You will notice this problem most readily in locales where Monday (or some other day) is the first day of the week. To test, invoke “SET DATEFIRST 1″ and try the provided function.
To get past this issue, you can try the following:
(DATEPART(dw, @SomeDate) + @@DATEFIRST) % 7
When you do this, you’ll see that a Wednesday is 4, regardless of the DATEFIRST setting.
Note that in this code the day is from 0 to 6, not 1 to 7.
i.e. saturday is 0.
1.how to get only system TIME using query
2.how to get only system DATE using query
Hi! Pinal,
I need to know, How can I display Date like
Mon, 20th Dec 2007
dbo.udf_PreviousThursday
=====================
Here is a useful UDF which takes a datetime value as input and returns as datetime output the previous Thursday:
ALTER FUNCTION dbo.udf_PreviousThursday(@dtDate DATETIME)
RETURNS DATETIME
AS
BEGIN
DECLARE @dt_PreviousThursday DATETIME
DECLARE @i_SubtractBy int –the number of days you must subtract from the input date to get the previous Thursday
SELECT @i_SubtractBy =
CASE DATEPART(weekday,@dtDate)
WHEN 1 THEN -4 –(i.e when Monday subtract 4 days to get the previous Thursday)
WHEN 2 THEN -5 –(i.e when Tuesday subtract 5 days to get the previous Thursday)
WHEN 3 THEN -6 –(i.e when Wednesday subtract 6 days to get the previous Thursday)
WHEN 4 THEN -7 –(i.e when Thursday subtract 7 days to get the previous Thursday) should this be zero??
WHEN 5 THEN -1 –etc.
WHEN 6 THEN -2 –etc.
WHEN 7 THEN -3 –etc.
END
SELECT @dt_PreviousThursday = DATEADD(day, @i_SubtractBy, @dtDate)
RETURN @dt_PreviousThursday
END
GO
–Sample execute
SET DATEFIRST 1–set Monday i.e. not Sunday to be the first day of the week
SELECT dbo.udf_PreviousThursday(getdate())
SELECT dbo.udf_PreviousThursday(’2007-01-12 13:52:09.043′)
In reply to Smita’s question, one of the existing datetime conversion formats can normally be relied upon for the conversion of datetime to string
e.g.
SELECT CONVERT(VARCHAR(100), GETDATE(), 106)
A list of common conversion styles is here:
http://technet.microsoft.com/en-us/library/aa237895(SQL.80).aspx
Of course, if you want to do something quite specific, you will have to use DATEPARTs to extract the relevant parts of the date, and concatenate these together. You need to convert all the parts to varchar prior to concatenation or you’ll get an error.
As a useful tip also, if you are getting 1/12/2007 and you want to get 01/12/2007 you need to use the right function like so:
right( ’00′ + Cast(datepart(dd, getdate()) as varchar), 2)
Hi pinal,
I need to know, How can I count the working days excluding sunday and saturday.
like if i need to count 7 working days..and today is monday so my seventh working days should be tuesday and i have to display the tuesday instead of the date fallin on tuesday.
Any help will be appreciated .
Regards
Vivek
@vivek
Give me some more details, like are you trying to retrieve data from a table, if yes then give the table structure what are the columns … that will be help me a little bit.
Thanks.
This function will let you correct day of the week considering the fact that DateFirst need not be always equal to 1
CREATE FUNCTION dbo.udf_DayOfWeek(@dtDate DATETIME)
RETURNS VARCHAR(10)
AS
BEGIN
DECLARE @rtDayofWeek VARCHAR(10)
DECLARE @weekDay int
– Here I have subtracted 7 For keeping Sunday as the First day
– like wise for Monday we need to subtract 2 and so on
@weekDay = Select ((DatePart(dw,GETDATE())+@@DATEFIRST-7)%7)
SELECT @rtDayofWeek = CASE @weekDay
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
RETURN (@rtDayofWeek)
END
GO
[...] 23, 2008 by pinaldave I have written article about SQL SERVER – UDF – Get the Day of the Week Function. I have received good modified script from reader Mihir Popat has suggested another code where [...]
Everything shows as Monday, regardless of the date. UGH
Hi, I want to get the dates of working days in a week
Can anybody help me.
Thanks in advance,
Partha
Dear Sir,
I need your help. I want to get the starting date and ending date for input week number. I am using sql server 2005.
Please do reply me soon.
Byyee and have a nice day.
Sunil,
Good question, I need the same solution.
Chris
Hi,
Please find this code..it might helps you.
–it calculates based upon current years Week#
DECLARE @week INT,
@weekdt DATETIME
SELECT @week = 40 –For Current Week
–get First day of current year
SELECT @weekdt = DATEADD(wk, @week, DATEADD(yy, DATEDIFF(yy,0,getdate()), 0))
SELECT @week AS Week#,
DATEADD(DD, 1 – DATEPART(DW, @weekdt), @weekdt) As WeekStartDate,
DATEADD(DD, 6, DATEADD(DD, 1 – DATEPART(DW, @weekdt), @weekdt)) As WeekEndDate
Can i get spanish names whit DATENAME ?
Yes Only if the default language of the server is Spanish
my question is i putup a any date of future and want to make calendre according to this way of any flight
my table is like
flightno sun mon tues wed thurs friday saturday
1001 y y n n n y n
1002 n y n y y n y
Sebastian,
This may help you
;WITH CTE AS
(
SELECT
1 AS ID,
SUBSTRING(days,0,CHARINDEX(‘,’,days)) AS Days,
SUBSTRING(days,CHARINDEX(‘,’,days)+1,LEN(days))RemainingStr,
2%8 AS DayID
FROM
sys.sysLanguages
WHERE
alias = ‘spanish’
UNION ALL
SELECT
ID+1 AS ID ,
CASE WHEN CHARINDEX(‘,’,RemainingStr) > 0 THEN SUBSTRING(RemainingStr,0,CHARINDEX(‘,’,RemainingStr))ELSE RemainingStr END AS Days,
SUBSTRING(RemainingStr,CHARINDEX(‘,’,RemainingStr)+1,LEN(RemainingStr))RemainingStr,
(DayID+1)%8 AS DayID
FROM
CTE
WHERE
ID < 7
)
SELECT Days,DAyID FROM CTE
HI, great post…
But, it’s possible with only data result “15-07-2007″ (called date) you know what is the first day of the week?
For example I want this result:
datepart(month, date) + ‘ ‘ + datepart(week, date) + ‘ ‘ + “First day of the week”
How I catch “First day of the week” ???
thanks in advance
Probably you need
select dateadd(week,datediff(week,0,date),0)
Time and Date only
SELECT CONVERT(VARCHAR(12), GETDATE(), 106)
SELECT CONVERT(VARCHAR(8), GETDATE(), 108)
@Nagarajan,
Note that all your TIME and DATEs are just VARCHARs
You should do this formation in the front end application
or
If you use SQL Server 2008, make use of new datatypes Time and Date
Hi All,
I want to show the first date of the week, example today date 25-02-2010, i want to 21-02-2010.
Inputs given year,month,week.(example:2010,02,4)
i want to output of this week first date
Gururajan.K
Hi Pinal,
I want the start date of week and the end date of week.A date will be passed as a parameter,based on the week number of that date I need the start & end date of that week.
Hi,
Can anyone help me to get the first “Sunday” in (Integer) of the given date… which will return the day of the first sunday
Example:
Date: ’05/22/2010′
Then I want to get the day (integer) of the first sunday of this date ’05/22/2010′
It should Return 2
Because the first sunday in the month of May 2010 is 2
Continued…
Because I want to get the Week1 to Week5
In my example there are 5 sundays in the Month of May 2010
Week 1 – date will be May 1 to May 2
Week 2 – date will be May 3 to May 9
Week 3 – date will be May 10 to May 16
Week 4 – date will be May 17 to May 23
Week 5 – date will be May 24 to May 30
Week 6 – date will be May 31
The only thing i want to know is the first sunday of the given date in integer in this case…
I want to get the 2,
Because the first sunday in the month of May 2010 is 2
just want to ask…
how to parse the each date in a year into MS SQL database? if i must store those date in an array?? if so, how can i parse them ??
hope to hear some comments…..
really need.
how to display weeknumbers and and its correponding year?
Hi, sorry I’m new on SQL codification … could you please tell me how can I include this function in a SELECT codification?
Regards.
Alejandro
Hi Pinal,
i have day of week , on the basis of day of week i want to find the next particular day (date)
for Example i have 3 =Tuesday ,i want to calculate next tuesday date on basis of current current date
Hi,
I want Monday – Sunday week numbers.
I have to group by this week number and i have to get count of records in between that week.
Thanks in advance.
Hi,
I want a method like DATEPART(WK,getdate()) , but it should take from monday – sunday.
Thanks in advance.
HI Everyone,
Can please help me any one on this issue:
when i’m runing this query:
SELECT DATENAME(wk,GETDATE())
ans:33
but,with this :
SELECT DATENAME(wk,GETDATE()-1)
ans:33
can anyone plz tel why it is this ,actualy i think output should be 32 when we do -1.
thanks in advance
SELECT DATENAME(wk,GETDATE())-1
ans: 32
thanks
I want to subtract from StartDate from current Date
from Table
select Hours = datediff(hh,0,DtDiff,StartDate) , Minutes = datepart(minute,DtDiff)
(select DtDiff = convert(datetime,’20110715 10:5:45′)- convert(DateTime,GetDate())
)a
from
Table Name
this is my query plz correction on it
To get a deterministic value for the day of week for a given date you could use a combination of datepart and @@firstday. Otherwise your dependent on the settings on the server.
Check out the following site for a better solution:
http://www.lazerwire.com/2011/10/sql-day-of-week.html
Hi,
I have a question that, I want to see a Table design in SQL Server. So which query I want to write to see the the design.
Thanks and Regards,
Madhu…
EXEC sp_help ‘table_name’
Hi I have created computed col with used following function for GET WEEK OF THE DATE but when create a index on that it says this is Non deterministic function, please help me on that how can i convert it into deterministic
create function dev_log.F_ISO_WEEK_OF_YEAR
(
@Date datetime
)
returns int
as
/*
Function F_ISO_WEEK_OF_YEAR returns the
ISO 8601 week of the year for the date passed.
*/
begin
declare @WeekOfYear int
select
— Compute week of year as (days since start of year/7)+1
— Division by 7 gives whole weeks since start of year.
— Adding 1 starts week number at 1, instead of zero.
@WeekOfYear =
(datediff(dd,
— Case finds start of year
case
when NextYrStart <= @date
then NextYrStart
when CurrYrStart <= @date
then CurrYrStart
else PriorYrStart
end,@date)/7)+1
from
(
select
— First day of first week of prior year
PriorYrStart =
dateadd(dd,(datediff(dd,-53690,dateadd(yy,-1,aa.Jan4))/7)*7,-53690),
— First day of first week of current year
CurrYrStart =
dateadd(dd,(datediff(dd,-53690,aa.Jan4)/7)*7,-53690),
— First day of first week of next year
NextYrStart =
dateadd(dd,(datediff(dd,-53690,dateadd(yy,1,aa.Jan4))/7)*7,-53690)
from
(
select
–Find Jan 4 for the year of the input date
Jan4 =
dateadd(dd,3,dateadd(yy,datediff(yy,0,@date),0))
) aa
) a
return @WeekOfYear
end
go
i want to display week days dates of the month from created date .
E: if i select radio button as week days on 11/11/2011
it should calculate nov months week dates
if i select calender from to to date it should select only week end dates
nice blog thanks
SELECT DATENAME(weekday, ’1988/08/31′) will do
It doesn’t have to be anywhere as near as complicated as that try this:
TO_CHAR(column_name, ‘Day dd-mon-yy’)
you can include this in other scripts.
I used this in a report i run daily to let me know when invoices need to be authorised. This script excludes weekends in a CASE:
TO_CHAR (( CASE
WHEN TO_CHAR(ps.run_date,’DY’) = ‘MON’
THEN TRUNC(ps.run_date)-3
ELSE TRUNC(ps.run_date)-1
END), ‘Day dd-mon-yy’) AS AUTHORISE_BY
Oh yes, I’m English, so i spell authorised in an English way.
I want to those record for the start date are from Monday to Sunday in sql query
Thanks
Hi,
I want to create one attendance report so,in that i want two columns one is “Total Working days” of the week and second column is “number of days present”. please help me for the same.