Jump to content
Sign in to follow this  
ricky

_DateDayOfWeek output incorrect

Recommended Posts

Hello,

here is the description :

#include <Date.au3>

; ISO Week day number for a given date 0=Monday - 6=Sunday

$iWeekday = _DateToDayOfWeekISO (@YEAR, @MON, @MDAY)

; NOT equal to @Wday

MsgBox(4096, "", "Todays ISO WeekdayNumber is: " & $iWeekDay)

The 1=Monday - 7=Sunday, what is wrong, the function or the description?

Share this post


Link to post
Share on other sites

; #FUNCTION# ====================================================================================================================
; Name...........: _DateToDayOfWeek
; Description ...: Returns the weekdaynumber for a given date.
; Syntax.........: _DateToDayOfWeek($iYear, $iMonth, $iDay)
; Parameters ....: $iYear  - A valid year in format YYYY
;                 $iMonth - A valid month in format MM
;                 $iDay   - A valid day in format DD
; Return values .: Success - Returns Day of the Week Range is 1 to 7 where 1=Sunday.
;                 Failure - 0 and Set @ERROR to:
;                 |0 - No error.
;                 |1 - Invalid Input Date
; Author ........: Jos van der Zande <jdeb at="" autoitscript="" dot="" com="">
; Modified.......:
; Remarks .......:
; Related .......: _DateToDayOfWeekISO, _DateDayOfWeek, _DayValueToDate, _DateAdd, _DateDiff
; Link ..........:
; Example .......: Yes
; ===============================================================================================================================
Func _DateToDayOfWeek($iYear, $iMonth, $iDay)
    ; Verify If InputDate is valid
    If Not _DateIsValid($iYear & "/" & $iMonth & "/" & $iDay) Then
        Return SetError(1,0,"")
    EndIf
    Local $i_aFactor = Int((14 - $iMonth) / 12)
    Local $i_yFactor = $iYear - $i_aFactor
    Local $i_mFactor = $iMonth + (12 * $i_aFactor) - 2
    Local $i_dFactor = Mod($iDay + $i_yFactor + Int($i_yFactor / 4) - Int($i_yFactor / 100) + Int($i_yFactor / 400) + Int((31 * $i_mFactor) / 12), 7)
    Return ($i_dFactor + 1)
EndFunc   ;==>_DateToDayOfWeek

; #FUNCTION# ====================================================================================================================
; Name...........: _DateToDayOfWeekISO
; Description ...: Returns the ISO weekdaynumber for a given date.
; Syntax.........: _DateToDayOfWeekISO($iYear, $iMonth, $iDay)
; Parameters ....: $iYear  - A valid year in format YYYY
;                 $iMonth - A valid month in format MM
;                 $iDay   - A valid day in format DD
; Return values .: Success - Returns Day of the Week Range is 1 to 7 where 1=Monday.
;                 Failure - 0 and Set @ERROR to:
;                 |0 - No error.
;                 |1 - Invalid Input Date
; Author ........: Jos van der Zande <jdeb at="" autoitscript="" dot="" com="">
; Modified.......:
; Remarks .......:
; Related .......: _DateToDayOfWeek, _DateDayOfWeek, _DayValueToDate, _DateAdd, _DateDiff
; Link ..........:
; Example .......: Yes
; ===============================================================================================================================
Func _DateToDayOfWeekISO($iYear, $iMonth, $iDay)
    Local $idow = _DateToDayOfWeek($iYear, $iMonth, $iDay)
    If @error Then
        Return SetError(1,0,"")
    EndIf
    If $idow >= 2 Then Return $idow - 1
    Return 7
EndFunc   ;==>_DateToDayOfWeekISO

_DateToDayOfWeek supposedly returns: Su=1, Mo=2, Tu=3, Wo=4, Thu=5, Fr=6, Sat=7

_DateToDayOfWeekISO takes the above and if result >= 2 then return subtract 1. Else return 7.

Knowing the above. _DateToDayOfWeekISO should return: Su=7, Mo=1, Tu=2, Wo=3, Thu=4, Fr=5, Sat=6

The comment above the function says the same thing. In this case, I think the help is wrong. (Unless the intention was other then what was written in the function)

Edited by Manadar

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
Sign in to follow this  

  • Recently Browsing   0 members

    No registered users viewing this page.

×
×
  • Create New...