@WDAY with NOT current date

[mireazma]

Hello.

I searched for the use of @WDAY in the way I want to but I only found it related to current date. I don't know how macros work and what they return, especially @WDAY. The manual is ambiguous. Would somebody please guide me how to use it to enumerate the dates from a starting one to an ending one, counting only the workdays?

Here is something I'm trying to make, more or less pseudo-coded:

; arbitrary month/year
$StartDate = 1
$EndDate = 14

For $Date = StartDate To EndDate
   If (@WDAY($Date) <> 1 OR @WDAY($Date) <> 7) Then
     <write the date output>
   EndIf
Next

Now, I've got the feeling $Date must be formatted somehow and I don't know how to use @WDAY. Besides, am I missing anything else?

[jchd]

_DateToDayOfWeek, DateToDayOfWeekISO

[sahsanu]

As jchd pointed out to you... just an example:

#include <Date.au3>
$StartDate = 1
$EndDate = 40 ;I know I know ;-)

For $Date = $StartDate To $EndDate
   $result = _DateToDayOfWeek (@YEAR, @MON,$Date)
   If Not @error Then
        If $result <> 1 And $result <> 7 Then
            ConsoleWrite(@YEAR & "/" & @MON & "/" & $Date & @TAB & _DateDayOfWeek($result) & @CRLF)
        EndIf
    EndIf
Next

[mireazma]

Thank you very much. (I always hated that English has the same form for 2nd person sg and pl. Now I meant both).

sahsanu, I don't think I could've done it without the snippet. It's unclear for me whether $Date must be string or numeric.

Edited May 4, 2011 by mireazma

[jchd]

Help file tells you what type functions expect/return.

[mireazma]

Yeah, it says "$iDay" - a valid day in format DD.

Accustomed to the rigorous C++, I find the variant being a source of ambiguity for me which not only I don't understand totally but as it seems it's a source of bugs as well. I couldn't test it at that moment so I couldn't know if I have to put "" around the assigned numerical constant. Now I know it has to be declared as string, otherwise it won't work.

Another example is @YEAR which I'm not sure what it returns. In the manual it says "Current four-digit year". Now is it a string or a number? Oh, I forgot it's both at the same time . Takes time to get accustomed to variants

[jchd]

ConsoleWrite(VarGetType(@YEAR) & @LF)

String

This is to adhere to the YYYY format and common stringed dates.

[mireazma]

Well, I learned something new - VarGetType. I didn't have the time to read althrough the manual. I'm starting to think that anything you could imagine has a corresponding function in AutoIt (everything is covered)

[PsaltyDS]

The native AutoIt functions and operators will attempt to recast parameters to correct type on the fly. The macro @YEAR will return a string but you can still get away with this:

$iNext = @YEAR + 1
ConsoleWrite("Next year will be: " & $iNext & "; Type = " & VarGetType($iNext) & @LF)

Because of the arithmetic operation, AutoIt performs as if the first line was:

$iNext = Number(@YEAR) + 1

Note there is no guarantee that a particular UDF will do that for you.

Edited May 4, 2011 by PsaltyDS

[mireazma]

Thanks.