Function to convert column number to letter?

Question

Does anyone have an Excel VBA function which can return the column letter(s) from a number?

For example, entering 100 should return CV.

Answer 1

And a solution using recursion:

Function ColumnNumberToLetter(iCol As Long) As String

    Dim lAlpha As Long
    Dim lRemainder As Long

    If iCol <= 26 Then
        ColumnNumberToLetter = Chr(iCol + 64)
    Else
        lRemainder = iCol Mod 26
        lAlpha = Int(iCol / 26)
        If lRemainder = 0 Then
            lRemainder = 26
            lAlpha = lAlpha - 1
        End If
        ColumnNumberToLetter = ColumnNumberToLetter(lAlpha) & Chr(lRemainder + 64)
    End If

End Function

Answer 2

Furthering on brettdj answer, here is to make the input of column number optional. If the column number input is omitted, the function returns the column letter of the cell that calls to the function. I know this can also be achieved using merely ColumnLetter(COLUMN()), but i thought it'd be nice if it can cleverly understand so.

Public Function ColumnLetter(Optional ColumnNumber As Long = 0) As String
    If ColumnNumber = 0 Then
        ColumnLetter = Split(Application.Caller.Address(True, False, xlA1), "$")(0)
    Else
        ColumnLetter = Split(Cells(1, ColumnNumber).Address(True, False, xlA1), "$")(0)
    End If
End Function

The trade off of this function is that it would be very very slightly slower than brettdj's answer because of the IF test. But this could be felt if the function is repeatedly used for very large amount of times.

Answer 3

what about just converting to the ascii number and using Chr() to convert back to a letter?

col_letter = Chr(Selection.Column + 96)

Answer 4

This is a version of robartsd's answer (with the flavor of Jan Wijninckx's one line solution), using recursion instead of a loop.

Public Function ColumnLetter(Column As Integer) As String
    If Column < 1 Then Exit Function
    ColumnLetter = ColumnLetter(Int((Column - 1) / 26)) & Chr(((Column - 1) Mod 26) + Asc("A"))
End Function

I've tested this with the following inputs:

1   => "A"
26  => "Z"
27  => "AA"
51  => "AY"
702 => "ZZ"
703 => "AAA" 
-1  => ""
-234=> ""

Answer 5

Column letter from column number can be extracted using formula by following steps
1. Calculate the column address using ADDRESS formula
2. Extract the column letter using MID and FIND function

Example:
1. ADDRESS(1000,1000,1)
results $ALL$1000
2. =MID(F15,2,FIND("$",F15,2)-2)
results ALL asuming F15 contains result of step 1

In one go we can write
MID(ADDRESS(1000,1000,1),2,FIND("$",ADDRESS(1000,1000,1),2)-2)

Answer 6

Here's another way:

{

      Sub find_test2()

            alpha_col = "A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,W,Z" 
            MsgBox Split(alpha_col, ",")(ActiveCell.Column - 1) 

      End Sub

}