VB6 - A function for obtaining the Day of the Week given the Date

Alright, actually it's two functions; DayOfTheWeek_S and DayOfTheWeek_I.
DayOfTheWeek_S returns a string value with the day of the week, and DayOfTheWeek_I returns an integer value (0 to 6) of the same.

Perfect for a module....

Code:

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Option Explicit

Dim intYr As Integer
Dim intMo As Integer
Dim intDay As Integer

Public Function DayOfTheWeek_S(ByVal TheDate As Date) As String
 Select Case DayOfTheWeek_I(TheDate)
  Case 0
  DayOfTheWeek_S = "Sunday"
  Case 1
  DayOfTheWeek_S = "Monday"
  Case 2
  DayOfTheWeek_S = "Tuesday"
  Case 3
  DayOfTheWeek_S = "Wednesday"
  Case 4
  DayOfTheWeek_S = "Thursday"
  Case 5
  DayOfTheWeek_S = "Friday"
  Case 6
  DayOfTheWeek_S = "Saturday"
  Case Else
  DayOfTheWeek_S = "HOLY ****!"
 End Select
End Function

Public Function DayOfTheWeek_I(TheDate As Date) As Integer
 intYr = Year(TheDate)
 intMo = Month(TheDate)
 intDay = Day(TheDate)

 'Debug.Print intYr; "intYr"
 'Debug.Print intMo; "intMo"
 'Debug.Print intDay; "intDay"
' W = D + Y + Year + M + C
' |  |  |    |    |  ^Century Offset Number (see Century_Offset Function)
' |  |  |    |    Month Offset Number (see MonthOffset Function)
' |  |  |    Present Year sans Century (i.e. if it's 2008, this number would be "8")
' |  |  'YearOffset...
' |  Day of the Month (i.e. if the Date is Janueary 1st, this number would be a "1")
' Day of the Week! (More or less;)
                         
 DayOfTheWeek_I = intDay + YearOffset + IntYear + MonthOffset + CenturyOffset
 DayOfTheWeek_I = DayOfTheWeek_I Mod 7
End Function

Private Function MonthOffset() As Integer
 Select Case intMo
  Case 1
  MonthOffset = 0
  If LeapYear = True Then MonthOffset = 6
  Case 2
  MonthOffset = 3
  If LeapYear = True Then MonthOffset = MonthOffset - 1
  Case 3
  MonthOffset = 3
  Case 4
  MonthOffset = 6
  Case 5
  MonthOffset = 1
  Case 6
  MonthOffset = 4
  Case 7
  MonthOffset = 6
  Case 8
  MonthOffset = 2
  Case 9
  MonthOffset = 5
  Case 10
  MonthOffset = 0
  Case 11
  MonthOffset = 3
  Case 12
  MonthOffset = 5
 End Select
 'Debug.Print MonthOffset; "MonthOffset"
End Function

Private Function LeapYear() As Boolean
 If IntYear = 0 Then 'Century!
  If Century / 400 = Century \ 400 Then LeapYear = True
  If Century / 4000 = Century \ 4000 Then LeapYear = False
 Else 'not round century...
  If (Century Mod 4) = 0 Then LeapYear = True
 End If
 'Debug.Print LeapYear; "LeapYear"
End Function

Private Function IntYear() As Integer
 IntYear = intYr - (IntCentury * 100)
 'Debug.Print IntYear; "IntYear"
End Function

Private Function IntCentury() As Integer
 IntCentury = intYr \ 100
 'Debug.Print IntCentury; "IntCentury"
End Function

Private Function CenturyOffset() As Integer
 CenturyOffset = 2 * (3 - (IntCentury Mod 4))
 'Debug.Print CenturyOffset; "CenturyOffset"
End Function

Private Function YearOffset() As Integer
 YearOffset = IntYear \ 4
 'Debug.Print YearOffset; "YearOffset"
End Function

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I'm sure this could be optimized, I only needed it to work. :D
I've gotten alot from this forum, glad nobody's posted this yet so I can give a little back.

Thanks for sharing :thumb:

That can actually be simplified quite a bit by making use of the Format function, which provides lots of functionality for converting Dates etc:

Code:

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Public Function DayOfTheWeek_S(ByVal TheDate As Date) As String
  DayOfTheWeek_S = Format(TheDate, "dddd")
End Function

Public Function DayOfTheWeek_I(TheDate As Date) As Integer
  DayOfTheWeek_I = Format(TheDate, "w") - 1
End Function

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(I added the -1 because the format function returns "1 for Sunday through 7 for Saturday", which is 1 higher than you wanted).

I echo si_the_geek's comments. Thanks for sharing. :thumb:

The leap year testing could also be optimized a bit by using the DateSerial function

vb Code:

Private Function IsThisALeapYear(ByVal intYear As Integer) As Boolean
     IsThisALeapYear = (29 = Day(DateSerial(intYear, 2, 29)))
End Function

Ah, gone and re-coded the wheel, have I?

Well, live and learn. Thanks for the instructional words of wisdom.

Bit late in the day but this way is even faster by using the Weekday function

Usage

Code:

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Dim Myday as integer
Dim MyDayS as String
MyDay = Weekday(Date) 'returns a number 1 .. 7 (Sunday ... Saturday)

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Then just do a few if's and Else's to convert the number to a day string ... such as

Code:

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If myday = 1 then
    MyDayS = "Sunday"
elseif Myday = 2 then
    MyDayS = "Monday"
etc.
etc.
Else
    MydayS = "Saturday"
End If

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Alternately you can use the Format Function to grab the day from the current, or any given date

Code:

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Dim MyStr As String
MyStr = Format(Date, "ddd")  ' Returns Day name ie. Mon, Tue, Wed etc

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