SET DATEFIRST in FUNCTION

Question

I want to SET DATEFIRST in my function but it is not allowed.

SET DATEFIRST 1

I can add the code in a SP and call the SP from the function but I am not keen on doing that.

I can SET the DATEFIRST before I call my function but I am not keen on doing that as well.

Any other work around?

EDIT

Below is the code I want to use in my FUNCTION to return the total working days of the month. But I cant add this code into the FUNCTION because of my DATEFIRST

DECLARE @my int
DECLARE @myDeduct int
DECLARE @day INT
DECLARE @mydate DATETIME
DECLARE @TotalDays INT

SET @mydate = GETDATE()

SET @myDeduct = 0
IF (@@DATEFIRST + DATEPART(DW, @mydate)) % 7 not in (0,1)
SET DateFirst 1 -- Set it monday=1 (value)

--Saturday and Sunday on the first and last day of a month will Deduct 1
IF (DATEPART(weekday,(DATEADD(dd,-(DAY(@mydate)-1),@mydate))) > 5)
SET @myDeduct = @myDeduct + 1

IF (DATEPART(weekday,(DATEADD(dd,-(DAY(DATEADD(mm,1,@mydate))),DATEADD(mm,1,@mydate)))) > 5)
SET @myDeduct = @myDeduct + 1

SET @my = day(DATEADD(dd,-(DAY(DATEADD(mm,1,@mydate))),DATEADD(mm,1,@mydate)))

Set @TotalDays = (select (((@my/7) * 5 + (@my%7)) - @myDeduct))

Select @TotalDays

Solution 1

My usual workaround is to use "known-good" dates for my comparisons.

Say, for instance, that I need to check that a date is a saturday. Rather than relying on DATEFIRST or language settings (for using DATENAME), I instead say:

DATEPART(weekday,DateToCheck) = DATEPART(weekday,'20120714')

I know that 14th July 2012 was a Saturday, so I've performed the check without relying on any external settings.

---

The expression (DATEPART(weekday,DateToCheck) + @@DATEFIRST) % 7 will always produce the value 0 for Saturday, 1 for Sunday, 2 for Monday, etc.

So, I'd advise you to create a table:

CREATE TABLE WorkingDays (
    NormalisedDay int not null,
    DaysInMonth int not null,
    WorkingDays int not null
)

Populating this table is a one off exercise. NormalisedDay would be the value computed by the expression I've given above.

To compute the DaysInMonth given a particular date, you can use the expression:

DATEDIFF(day,
      DATEADD(month,DATEDIFF(month,0,DateToCheck),0),
      DATEADD(month,DATEDIFF(month,'20010101',DateToCheck),'20010201'))

Now all your function has to do is look up the value in the table.

(Of course, all of the rows where DaysInMonth is 28 will have 20 as their result. It's only the rows for 29,30 and 31 which need a little work to produce)

Solution 2

Instead of

SET DATEFIRST 1

You can do

SELECT (DATEPART(weekday, GETDATE()) + @@DATEFIRST - 2) % 7 + 1

Comments

[Mikael Eriksson]

Perhaps you could rework the logic in your function so you are not dependent on set datefirst?

[tenfour]

What is your real problem, not your proposed solution?

[Etienne]

Please see above, I added my code in to explain how I want to use it.

[Etienne]

Please see above, I added my code in to explain how I want to use it.

[Damien_The_Unbeliever]

@Etienne - I've added some useful expressions and a proposed solution.

[underscore_d]

+1 for (DATEPART(weekday,DateToCheck) + @@DATEFIRST) % 7, invaluable! sure beats comparing the weekday names, anyway :O

[ArieKanarie]

This is a good one, nice one liner. Gives you 1 for Monday, 7 for Sunday. Needed this for ISO8601 calculations. See stackoverflow.com/a/41248745/2997016 for a multi line version if modulo is like abracadabra for you

[underscore_d]

IMO this doesn't really the question, as it still depends on the current value of @@datefirst. It just 'rotates' the weekdays based on that, so you will only get 1 for Monday if the current @@datefirst is Sunday. That may suffice for some readers, but it's not a comprehensive answer like Kvasi's. As another note, isn't the <= 7 redundant, because there cannot be > 7 weekdays in a week?

[underscore_d]

@ArieKanarie but the answer you linked, unlike this one, doesn't account for the current value of @@datefirst, so it only works if that was such that Sunday is weekday 1. Maybe you could post another answer that fixes that. :P

[underscore_d]

Anyway, +1, though I used a modified version without the subtraction and addition, since I only need the numbers to be consistent, not specific. So 0 is Saturday, 1 is Sunday, etc. This means I (A) can accommodate datefirst in a function and (B) without the region-specific, slower kludge of comparing to specific datenames instead.

[ArieKanarie]

@underscore_d Yes the <= 7 could also be a simple ELSE. At least until we get 8 days a week ;-)