Day of Year

This code has been declared by its author to be Public Domain code.

Day of Year by xlsior

2007

The code below will return the day of the year for any date, taking leap years into consideration as well.

e.g. January 1st will be 1, December 31st will be either 365 or 366, depending on leap years.

http://www.xlsior.org

' Day of the Year
'
' By Marc van den Dikkenberg
'
Strict
Print "Day of the year: "+DayNumber("05 Aug 2007")	' Specific Date
Print "Day of the year: "+DayNumber("")					' Returns Current Date


Function DayNumber(SomeDate:String)		' Date in form: "DD MMM YYYY", e.g.  "02 APR 2007"
	If SomeDate="" Then	
		SomeDate = CurrentDate$() 		' If no date specified, use today's date
	End If

	Local Dag:Int[]=[31,28,31,30,31,30,31,31,30,31,30,31,31,29,31,30,31,30,31,31,30,31,30,31]
	Local Maand:String[]=["JAN","FEB","MAR","APR","MAY","JUN","JUL","AUG","SEP","OCT","NOV","DEC"]
	Local YearStartPoint:Int=0
	Local Jaar:Int=Int(Right$(SomeDate,4))
	Local MonthCount:Int=0
	Local TotalDays:Int=0
	Local TempVar:Int=0

	If ((Jaar Mod 4) = 0 And (Jaar Mod 100) <> 0) Or ((Jaar Mod 4) = 0 And (Jaar Mod 400) = 0) Then 
		' It's a Leap Year
		YearStartPoint=12
	Else
		' It's not a Leap Year
		YearStartPoint=0
	End If

	For TempVar:Int=0 To 11
		If Upper(Mid$(SomeDate,4,3))=maand[TempVar] Then
			MonthCount:Int=TempVar:Int
			Exit
		End If
	Next
	For Tempvar:Int=0 To MonthCount:Int-1
		TotalDays:Int=TotalDays:Int+dag[TempVar+YearStartPoint]
	Next
	TotalDays:Int=TotalDays:Int+Int(Left$(SomeDate,2))
	Return TotalDays
End Function

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