How to convert Julian date ignoring the year to Julian Day - vba

Question

How do i convert day/Month to Julian Day in visual basic?
Formula for converting in other languages are also appreciated.
For e.g
Julian Day for 1 Feb=032
https://landweb.modaps.eosdis.nasa.gov/browse/calendar.html
From research in the web ,most resolutions are converting the date with yyyy-mm-dd to julian date.
As the example above, i need to get value 032

Answer 1

The .Net JulianCalendar Class exposes all methods needed to achieve your goal. It is also COM visible, so you can reference it in a VBA project after adding a project reference to "mscorlib.dll".
In VB.Net the code would be:
Dim jc As New System.Globalization.JulianCalendar
' create a gregorian equivalent of the julian date: 1 Feb 2018
Dim gregoriaDateEquivalent As DateTime = jc.ToDateTime(2018, 2, 1, 0, 0, 0, 0)
' = #2/14/2018 12:00:00 AM#
Dim dayOfYear As Int32 = jc.GetDayOfYear(gregoriaDateEquivalent)
' = 32
In VBA the code would be:
Dim jc As mscorlib.JulianCalendar
Set jc = New mscorlib.JulianCalendar
' create a gregorian equivalent of the julian date: 1 Feb 2018
Dim gregoriaDateEquivalent As Date
gregoriaDateEquivalent = jc.ToDateTime(2018, 2, 1, 0, 0, 0, 0)
' = #2/14/2018
Dim dayOfYear As Long
dayOfYear = jc.GetDayOfYear(gregoriaDateEquivalent)
' = 32

Answer 2

First you have to add reference to mscorlib library
Under VB6,Project-References->Tick the mscorlib checkbox
Create this function method in the object that you ahve or otherwise create a sub procedure
Public Function JulianConverter() As String
'MsgBox Fix(5.2), vbExclamation, App.Title
'MsgBox Fix(5.6)
Dim y As String
Dim m As String
Dim d As String
Dim strDate As String
Dim dateArray() As String
strDate = Format(Now, "dd/MMM/yyyy")
dateArray = Split(strDate, "/")
d = dateArray(0)
m = dateArray(1)
y = dateArray(2)
'Debug
'MsgBox strDate
'Convert to Integer
y = Val(y)
m = Val(m)
d = Val(d)
If m <= 2 Then
y = y - 1
m = m + 12
End If
'Dim A As Double
' Dim B As Double
'Dim JD As Double
' A = CDbl(Fix(y / 100#))
'B = 2 - A + Fix(A / 4)
'JD = Fix(365.25 * (y + 4716)) + Fix(30.6001 * (m + 1)) + d + B - 1524.5
'JulianConverter = CStr(JD) 'Convert to string
Dim jc As mscorlib.JulianCalendar
Set jc = New mscorlib.JulianCalendar
' create a gregorian equivalent of the julian date: 1 Feb 2018
Dim gregoriaDateEquivalent As Date
gregoriaDateEquivalent = jc.ToDateTime(2018, m, d, 0, 0, 0, 0)
' = #2/14/2018
Dim dayOfYear As Long
Dim dayOfYearS As String
Dim digitLength As Integer
Dim Counter As Integer
dayOfYear = jc.GetDayOfYear(gregoriaDateEquivalent)
'Have to ensure 3 digits values
dayOfYearS = CStr(dayOfYear)
'Count number of Digits of string
digitLength = Len(dayOfYearS)
MsgBox "DigitLength" & digitLength, vbExclamation, App.Title
'If Digits length smaller than 3,add one 0 in front
If (digitLength < 3) Then
dayOfYearS = "0" & dayOfYearS
End If
JulianConverter = dayOfYearS
End Function
This will ensure the julian day will be 3 digits in value based on the current system date