Comments on: Column Letters To Numbers

By: Superplanilla

Superplanilla — Tue, 08 Jun 2004 00:56:00 +0000

Otra variante:

Function colnum(x)
colnum = Range(x & ":" & x).Column
End Function

By: Gary Waters

Gary Waters — Mon, 07 Jun 2004 23:55:00 +0000

Dick, here's a function that includes error checking and covers all positive long integers:
Function ColumnNumber(ByVal strCol As String) As Long
Dim length As Long, i As Long
strCol = UCase(strCol)
length = Len(strCol)
If length < 1 Or length > 7 Then
ColumnNumber = -1
Exit Function
End If
If Not strCol Like Replace(Space(length), " ", "[A-Z]") Then
ColumnNumber = -1
Exit Function
End If
If length = 7 Then
If strCol > "FXSHRXW" Then
ColumnNumber = -1
Exit Function
End If
End If
ColumnNumber = 0
For i = 1 To length
ColumnNumber = ColumnNumber + (Asc(Mid$(strCol, i, 1)) - 64) * 26 ^ (length - i)
Next i
End Function

By: Ron de Bruin

Ron de Bruin — Mon, 07 Jun 2004 20:32:00 +0000

Try this

Also from Chip I believe

Function ColumnNumber(ColLetter) As Integer
ColumnNumber = Cells(1, ColLetter).Column
End Function

Temp = ColumnNumber("D") ' returns 4

By: tjs

tjs — Mon, 07 Jun 2004 18:51:00 +0000

David,
this eliminates the extra math. i.e. (26 * (LenColLetter - i)

Function ColNum(ByVal ColumnLetter As String) As Long
Dim LenColLetter As Integer, i As Integer
ColNum = 0: LenColLetter = Len(ColumnLetter)
For i = 1 To LenColLetter
ColNum = ColNum * 26 + (Asc(UCase(Mid(ColumnLetter, i, 1))) - 64)
Next i
End Function

By: Dick

Dick — Mon, 07 Jun 2004 18:35:00 +0000

David: I think you need to get rid of the +1. ?ColNum(“a”) = 2.

By: Harald Staff

Harald Staff — Mon, 07 Jun 2004 18:30:00 +0000

David

THis is good. But I get “2? as a result of “A”. After removing the +1 it said 1. I ran this, combining the two:

Sub test()
Dim L As Long, L2 As Long, S As String
L = 0
Do
L = L + 1
S = ColumnLetter(L)
L2 = colNum(S)
If L2 <> L Then
MsgBox L & ” vs ” & L2, , S
Exit Sub
End If
Loop Until L > 1000000
MsgBox S, , L
End Sub

No error. So either both are great or both are equally wrong. Unfortunately I have only 256 columns to check against

Best wishes Harald

By: David Wasserman

David Wasserman — Mon, 07 Jun 2004 17:03:00 +0000

Oops, the 5th line should read:

ColNum = ColNum + (26 ^(LenColLetter - i) * (Asc(UCase(Mid(ColumnLetter, i, 1))) - 64)) + 1

The difference is power vs multiplication.

By: David Wasserman

David Wasserman — Mon, 07 Jun 2004 16:52:00 +0000

This function will work even if Microsoft expands the columns to allow three or more letters:

Function ColNum(ByVal ColumnLetter As String) As Long
Dim LenColLetter As Integer, i As Integer
ColNum = 0: LenColLetter = Len(ColumnLetter)
For i = 1 To LenColLetter
ColNum = ColNum + (26 * (LenColLetter - i) * (Asc(UCase(Mid(ColumnLetter, i, 1))) - 64)) + 1
Next i
End Function

By: CJM

CJM — Mon, 07 Jun 2004 16:25:00 +0000

Apologies for the formatting, the indents seem to fall out when it gets posted...

=======================================

Function ColNum(ByVal ColumnLetter As String) As Long
'Returns the column reference if valid. If reference is
'invalid, returns 0.
ColNum = 0
If Len(ColumnLetter) = 2 Then
If ValidColumn(Left(ColumnLetter, 1), False) And _
ValidColumn(Right(ColumnLetter, 1), True) Then
ColNum = ThisWorkbook.Sheets(1).Range(ColumnLetter & "1?).Column
Else
ColNum = 0
End If
ElseIf Len(ColumnLetter) = 1 Then
If ValidColumn(Left(ColumnLetter, 1), False) Then
ColNum = ThisWorkbook.Sheets(1).Range(ColumnLetter & "1?).Column
Else
ColNum = 0
End If
End If
End Function

Function ValidColumn(TestChar As String, IVTest As Boolean) As Boolean
ValidColumn = False
If IVTest Then
If Asc(UCase(TestChar)) >= 65 And Asc(UCase(TestChar)) < = 86 Then
ValidColumn = True
End If
Else
If Asc(UCase(TestChar)) >= 65 And Asc(UCase(TestChar)) <= 90 Then
ValidColumn = True
End If
End If
End Function

By: Andy Pope

Andy Pope — Mon, 07 Jun 2004 16:09:00 +0000

How about,

Function ColNum(ByVal ColumnLetter As String) As Long
On Error Resume Next
ColNum = Range(ColumnLetter & "1?).Column
End Function