Function getColNameFromIndex(ByVal colIndex As Integer) As String

getColNameFromIndex = “”

getColNameFromIndex = Left(Cells(1, colIndex).Address(1, 0), InStr(1, Cells(1, colIndex).Address(1, 0), “$”) – 1)

End Function

Function getColNameFromIndex(ByVal colIndex As Integer) As String

getColNameFromIndex = “”

If (colIndex = 16385) Then
Exit Function
End If

getColNameFromIndex = Left(Cells(1, colIndex).Address(1, 0), InStr(1, Cells(1, colIndex).Address(1, 0), “$”) – 1)

End Function

Function Alpha_Column(Cell_Add As Range) As String
Dim No_of_Rows As Integer
Dim No_of_Cols As Integer
Dim Num_Column As Integer
Dim firstColumnLetter As String
Dim secondColumnLetter As String

No_of_Rows = Cell_Add.Rows.Count
No_of_Cols = Cell_Add.Columns.Count

If ((No_of_Rows 1) Or (No_of_Cols 1)) Then
Alpha_Column = “”
Exit Function
End If
Num_Column = Cell_Add.Column
If Num_Column < 26 Then
Alpha_Column = Chr(64 + Num_Column)
Else
If Num_Column Mod 26 0 Then
firstColumnLetter = Chr(Int(Num_Column / 26) + 64)
secondColumnLetter = Chr((Num_Column Mod 26) + 64)
Else
firstColumnLetter = Chr(Int(Num_Column / 26) + 63)
secondColumnLetter = “Z”
End If

Alpha_Column = firstColumnLetter & secondColumnLetter
End If
End Function

Four characters shorter, but more importantly, two function calls less…

=LEFT(ADDRESS(ROW(),COLUMN(),2),FIND(“$”,ADDRESS(ROW(),COLUMN(),2))-1)

Just wanted to point out one major flaw with the original post. If the Num_Column is equal to 26 then the remainder will be 0. The value falls in the else condition where the first part of Alpha_Column formula “Chr(Int(Num_Column / 26) + 64)” is 65 which is equal to the ASCII Chr of A. In second part of the Alpha_Column formula “Chr((Num_Column Mod 26) + 64)” is the remainder, 0, plus 64, which is 64. The ASCII value for 64 is “@”. Alpha_Column is then equal to “A@” which is definitely wrong. So to correct this simply modify the operator in the statement, “If Num_Column < 26 Then" to "If Num_Column <= 26 Then".

returns column name correctly

[range object].column

where [range object] can be Selection, ActiveCell, or any variable declared as Range. Then use Cells(x, y) for whatever you need.

It’s easier for loops to use numbers than letters. (That’s meant to invoke a smile or two from those who get it.)

Two points… First, you don’t really need the row to be specified as you can use hard code any row number into the Cells property call (I would suggest using 1) since the column letter is not dependent on the row number. Second, you code will return the wrong answer if the column number is greater than 26.

Sub GetColumn(ByVal row As Integer, ByVal col As Integer, ByRef letter As String)
addr = Sheet1.Cells(row, col).Address(False, False)
letter = Left(addr, 1)
End Sub

Called like this:

Call GetColumn(r, c, colLetter)

Doing that stores the column letter in the variable colLetter. Integers can be input for r and c to indicate which cell you are talking about.