The following converts the version number, including lettered versions, to dotted decimal format for easy comparison. 4.6.3b becomes version 4.6302, which is definitely a "higher" version than 4.6.2a (4.6201 in this code).

' from Scott Purl (spurl@acm.org) ' Yep. Only one Dim. Session.NotesVersion holds the key. Dim Session As New NotesSession ' Find out the version information MyVersion = Session.NotesVersion ' Get the length of the version string Length = Len(MyVersion) ' Discover What the "half" delimiter is, meaning what's the first non-alphanumeric character starting from the left ' Note that in the North American and International versions, this first non-alpha character differs If MyVersion Like "*Intl*" Then SepChar = "(" OffSet = 0 Else SepChar = "|" OffSet = 2 End If ' Now find this first non-alphanumeric character SepLoc = Instr(MyVersion, SepChar) ' Get everything to the left of that character LeftHalf = Left(MyVersion, SepLoc - 1) ' Find the only space in the left half SpacLoc = Instr(LeftHalf, " ") ' The version number is what's left to the right of the space in the left half, minus extra spaces ' code (with bug) was VerNum = Trim(Right(LeftHalf, SpacLoc - OffSet)) VerNum = Trim(Right(LeftHalf, Len(LeftHalf) - SpacLoc - OffSet)) ' Get the length of the version number VerLen = Len(VerNum) ' Loop through the version number to remove periods For x =1 To VerLen CurChar = Mid(VerNum, x, 1) If CurChar = "." Then ' Do Nothing Else ' Keep appending non-period characters until finished SubVer = SubVer & CurChar End If Next ' If the right-most character is a letter, convert it to a number ' "a" becomes "01", and "z" becomes "26" RSubVer = Right(SubVer, 1) If Instr("abcdefghijklmnopqrstuvwxyz", RSubVer) Then ' The first 9 letters of the alphabet need a padding leading zero If Instr("abcdefghi", RSubVer) Then AlphaNum = Instr("abcdefghi", RSubVer) NewSubVer = Left(SubVer, Len(SubVer) -1) & "0" & AlphaNum End If ' The rest of the alphabet are all greater than 10 in position If Instr("jklmnopqrstuvwxyz", RSubVer) Then AlphaNum = Instr("abcdefghijklmnopqrstuvwxyz", RSubVer) NewSubVer = Left(SubVer, Len(SubVer) -1) & AlphaNum End If ' And the numeric equivalent of the position is SubVer = NewSubVer End If ' Assemble it into leftmost character, plus a period, plus the rest of the iterated sub version TextVer = Left(SubVer, 1) & "." & Right(SubVer, Len(Subver) -1) ' Convert it to a normal number so we can do quick comparisons. FinVer = Cstr(TextVer)

You can then do:

If FinVer >= 4.6302 Then 'Action if True Else 'Action if False End If