4 replies to this topic

[Andi]

I want to retrieve the date of the last day of the current month.

e.g.
31.01.05
28.02.05
31.03.05
etc.

Perhaps one could take the first day of the following month and subtract one day, but I don't know how to do this? Has someone a tip for me?

[daonlyfreez]

lastDayOfThisMonth =

30daysMonths = 04,06,09,11

leapYears = 08,12,16,20

;

formattime, currentMonth,%a_now%, MM

formattime, currentYear,%a_now%, YY

;

IfEqual, currentMonth, 02

{	

	if currentYear in %leapYears%

	{

		lastDayOfThisMonth = 29

	}

	else

	{

		lastDayOfThisMonth = 28

	}

}

else

{

	if currentmonth in %30daysMonths%

	{

		lastDayOfThisMonth = 30

	}

	else

	{

		lastDayOfThisMonth =31

	}

}

msgbox, The last day of this month is the %lastDayOfThisMonth%(th/st)

[Chris]

Good solution. I like date challenges so here's another one that seems correct:

if A_MM = 12

{

	NextMonth = 01

	Year := A_YYYY + 1

}

else

{

	NextMonth := A_MM + 1

	NextMonth = 0%NextMonth%  ; Add leading zero.

	StringRight, NextMonth, NextMonth, 2   ; Remove inappropriate leading zero, if any.

	Year := A_YYYY

}

FirstDayOfNextMonth = %Year%%NextMonth%01

FirstDayOfNextMonth += -1, day  ; Subtract one day to yield last day of this month.

StringMid, LastDay, FirstDayOfNextMonth, 7, 2  ; Extract just the day.

MsgBox Last day of this month: %LastDay%

[daonlyfreez]

Wow, I didn't know you could do direct math on the datestring and even without some sort of identifier... 8)

[Andi]

That's the answer to this problem.

Thank you!!!!!!