VBA Finding the next column based on an input value

Question

In a program that I\'m trying to write now I take two columns of numbers and perform calculations on them. I don\'t know where these two columns are located until the user tell

Answer 1

Make use of the Asc() and Chr() functions in VBA, like so:

Dim first_Column As String
Dim second_Column As String

first_Column = Range("B2").Text
second_Column = Chr(Asc(first_Column) + 1)

The Asc(s) function returns the ASCII code (in integer, usually between 0 and 255) of the first character of a string "s".

The Chr(c) function returns a string containing the character which corresponds to the given code "c".

Upper case letters (A thru Z) are ASCII codes 65 thru 90. Just google ASCII for more detail.

NOTE: The above code will be fine so long as the first_Column is between "A" and "Y"; for columns "AA" etc., it will take a little more work, but Asc() and Chr() will still be the ticket to coding for that.

Answer 2

You were on the right track with Offset. Here is a test function that shows a couple different approaches to take with it:

Sub test()

Dim first_Column As String
Dim second_Column As String
Dim third_Column As String
Dim r As Range

    first_Column = Range("B2").Text
    second_Column = Range("B2").Offset(1, 0).Text
    third_Column = Range("B2").Offset(2, 0).Text
    Debug.Print first_Column, second_Column, third_Column

    Set r = Range("B2")
    first_Column = r.Text
    Set r = r.Offset(1, 0)
    second_Column = r.Text
    Set r = r.Offset(1, 0)
    third_Column = r.Text
    Debug.Print first_Column, second_Column, third_Column

End Sub

UPDATE: After re-reading your question I realize you were trying to do offsets based on a user-entered column letter. @rskar's answer will shift the column letter, but it will be a lot easier to work with the column number in code. For example:

Sub test()
Dim first_Col As Integer, second_Col As Integer
    first_Col = Cells(, Range("B2").Text).Column
    second_Col = first_Col + 1

    Cells.Columns(first_Col).Font.Bold = True
    Cells.Columns(second_Col).Font.Italic = True
End Sub

Answer 3

There are a few syntactical problems with @rskar's answer. However, it was helpful in producing a function that grabs a column "letter", based on an input column "letter" and a desired offset to the right:

Public Function GetNextCol(TheCol As String, OffsetRight As Integer) As String
    Dim TempCol1 As String
    Dim TempCol2 As String
    TempCol1 = Range(TheCol & "1").Address
    TempCol2 = Range(TempCol1).Offset(0, OffsetRight).Address(0, 0, xlA1)
    GetNextCol = Left(TempCol2, Len(TempCol2) - 1)
End Function

Answer 4

In light of the comments of others (and they all raised valid points), here is a much better solution to the problem, using Offset and Address:

Dim first_Column As String
Dim second_Column As String
Dim p As Integer

first_Column = Range("B2").Text

second_Column = _
    Range(first_Column + ":" + first_Column).Offset(0, 1).Address(0, 0, xlA1)
p = InStr(second_Column, ":")
second_Column = Left(second_Column, p - 1)

The above should work for any valid column name, "Z" and "AA" etc. included.

Answer 5

Here are two functions that will help you dealing with columns > "Z". They convert the textual form of a column to a column index (as a Long value) and vice versa:

Function ColTextToInt(ByVal col As String) As Long
    Dim c1 As String, c2 As String
    col = UCase(col) 'Make sure we are dealing with "A", not with "a"
    If Len(col) = 1 Then  'if "A" to "Z" is given, there is just one letter to decode
        ColTextToInt = Asc(col) - Asc("A") + 1
    ElseIf Len(col) = 2 Then
        c1 = Left(col, 1)  ' two letter columns: split to left and right letter
        c2 = Right(col, 1)
        ' calculate the column indexes from both letters  
        ColTextToInt = (Asc(c1) - Asc("A") + 1) * 26 + (Asc(c2) - Asc("A") + 1)
    Else
        ColTextToInt = 0
    End If
End Function

Function ColIntToText(ByVal col As Long) As String
    Dim i1 As Long, i2 As Long
    i1 = (col - 1) \ 26   ' col - 1 =i1*26+i2 : this calculates i1 and i2 from col 
    i2 = (col - 1) Mod 26
    ColIntToText = Chr(Asc("A") + i2)  ' if i1 is 0, this is the column from "A" to "Z"
    If i1 > 0 Then 'in this case, i1 represents the first letter of the two-letter columns
        ColIntToText = Chr(Asc("A") + i1 - 1) & ColIntToText ' add the first letter to the result
    End If
End Function

Now your problem can be solved easily, for example

newColumn = ColIntToText(ColTextToInt(oldColumn)+1)

EDITED accordingly to the remark of mwolfe02:

Of course, if you are not interested in the column names, but just want to get a range object of a specific cell in a given row right beneath a column given by the user, this code is "overkill". In this case, a simple

 Dim r as Range
 Dim row as long, oldColumn as String
 ' ... init row and oldColumn here ...

 Set r = mysheet.Range(oldColumn & row).Offset(0,1)
 ' now use r to manipulate the cell right to the original cell

will do it.