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Does anyone have an Excel VBA function which can return the column letter(s) from a number?
For example, entering 100 should return CV.
So I'm late to the party here, but I want to contribute another answer that no one else has addressed yet that doesn't involve arrays. You can do it with simple string manipulation.
Function ColLetter(Col_Index As Long) As String
Dim ColumnLetter As String
'Prevent errors; if you get back a number when expecting a letter,
' you know you did something wrong.
If Col_Index <= 0 Or Col_Index >= 16384 Then
ColLetter = 0
ColumnLetter = ThisWorkbook.Sheets(1).Cells(1, Col_Index).Address 'Address in $A$1 format
ColumnLetter = Mid(ColumnLetter, 2, InStr(2, ColumnLetter, "$") - 2) 'Extracts just the letter
ColLetter = ColumnLetter
After you have the input in the format $A$1, use the Mid function, start at position 2 to account for the first $, then you find where the second $ appears in the string using InStr, and then subtract 2 off to account for that starting position.
This gives you the benefit of being adaptable for the whole range of possible columns. Therefore, ColLetter(1) gives back "A", and ColLetter(16384) gives back "XFD", which is the last possible column for my Excel version.
MsgBox Columns( 9347 ).Address
To return ONLY the column letter(s): Split((Columns(Column Index).Address(,0)),":")(0)
MsgBox Split((Columns( 2734 ).Address(,0)),":")(0)
This formula will give the column based on a range (i.e., A1), where range is a single cell. If a multi-cell range is given it will return the top-left cell. Note, both cell references must be the same:
How it works:
CELL("property","range") returns a specific value of the range depending on the property used. In this case the cell address.
The address property returns a value $[col]$[row], i.e. A1 -> $A$1.
The MID function parses out the column value between the $ symbols.
robertsd's code is elegant, yet to make it future-proof, change the declaration of n to type long
In case you want a formula to avoid macro's, here is something that works up to column 702 inclusive
where A1 is the cell containing the column number to be converted to letters.
Just one more way to do this. Brettdj's answer made me think of this, but if you use this method you don't have to use a variant array, you can go directly to a string.
ColLtr = Cells(1, ColNum).Address(True, False)
ColLtr = Replace(ColLtr, "$1", "")
or can make it a little more compact with this
ColLtr = Replace(Cells(1, ColNum).Address(True, False), "$1", "")
Notice this does depend on you referencing row 1 in the cells object.
This is a version of robartsd's answer (with the flavor of Jan Wijninckx's one line solution), using recursion instead of a loop.
Public Function ColumnLetter(Column As Integer) As String
If Column < 1 Then Exit Function
ColumnLetter = ColumnLetter(Int((Column - 1) / 26)) & Chr(((Column - 1) Mod 26) + Asc("A"))
I've tested this with the following inputs:
1 => "A"
26 => "Z"
27 => "AA"
51 => "AY"
702 => "ZZ"
703 => "AAA"
-1 => ""
I've just noticed that this is essentially the same as Nikolay Ivanov's solution, which makes mine a little less novel. I'll leave it up because it shows a slightly different approach for a few of the minutia
This function returns the column letter for a given column number.
Function Col_Letter(lngCol As Long) As String
vArr = Split(Cells(1, lngCol).Address(True, False), "$")
Col_Letter = vArr(0)
testing code for column 100
You can add the (0) to the end of the Split command if you want to save yourself a variable declaration and extra line of code. eg Col_letter = Split(Cells(1, lngCol).Address(True, False), "$")(0)
Col_letter = Split(Cells(1, lngCol).Address(True, False), "$")(0)
That is quite correct, but I thought it more readable to use several lines.
Why bother with the Boolean params in this situation. You can do this:................................................... v = Split(Cells(1, lngCol).Address, "$")(1)
v = Split(Cells(1, lngCol).Address, "$")(1)
@MátéJuhász Not true. Split always returns a 0 based array, regardless of Option Base. Option Base specifies the default lower bound when its not otherwise specified
While this is very old, I have a minor addition - checking first if the number is positive, since otherwise you run into errors. if lngcol <=0 then
When using VBS, remember that .Cells is a property of Excel, meaning you need to use <excel_object>.Cells(). Otherwise, you will get a type mismatch error.
This is a function based on @DamienFennelly's answer above. If you give me a thumbs up, give him a thumbs up too! :P
Function outColLetterFromNumber(iCol as Integer) as String
sAddr = Cells(1, iCol).Address
aSplit = Split(sAddr, "$")
outColLetterFromNumber = aSplit(1)
Good one, but how is it different from the accepted answer?
@loannis I based mine on DamianFennelly's answer, not the accepted one. But yeah, mine looks a lot like the accepted answer, except one line is broken into two to make it more readable.
Here, a simple function in Pascal (Delphi).
function GetColLetterFromNum(Sheet : Variant; Col : Integer) : String;
Result := Sheet.Columns[Col].Address; // from Col=100 --> '$CV:$CV'
Result := Copy(Result, 2, Pos(':', Result) - 2);
Function fColLetter(iCol As Integer) As String
On Error GoTo errLabel
fColLetter = Split(Columns(lngCol).Address(, False), ":")(1)
fColLetter = "%ERR%"
what about just converting to the ascii number and using Chr() to convert back to a letter?
col_letter = Chr(Selection.Column + 96)
it doesn't work for column CV in the question:(
Here is a late answer, just for simplistic approach using Int() and If in case of 1-3 character columns:
Function outColLetterFromNumber(i As Integer) As String
If i < 27 Then 'one-letter
col = Chr(64 + i)
ElseIf i < 677 Then 'two-letter
col = Chr(64 + Int(i / 26)) & Chr(64 + i - (Int(i / 26) * 26))
col = Chr(64 + Int(i / 676)) & Chr(64 + Int(i - Int(i / 676) * 676) / 26)) & Chr(64 + i - (Int(i - Int(i / 676) * 676) / 26) * 26))
outColLetterFromNumber = col
Cap A is 65 so:
MsgBox Chr(ActiveCell.Column + 64)
Found in: http://www.vbaexpress.com/forum/showthread.php?6103-Solved-get-column-letter
this is only for REFEDIT ... generaly use uphere code
shortly version... easy to be read and understood /
it use poz of $
Private Sub RefEdit1_Change()
Me.Label1.Caption = NOtoLETTER(RefEdit1.Value) ' you may assign to a variable var=....'
Dim First As Long, Second As Long
First = InStr(REFedit, "$") 'first poz of $
Second = InStr(First + 1, REFedit, "$") 'second poz of $
NOtoLETTER = Mid(REFedit, First + 1, Second - First - 1) 'extract COLUMN LETTER
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), "$")(0)
ColumnLetter = Split(Cells(1, ColumnNumber).Address(True, False, xlA1), "$")(0)
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.
Dim Chara As String
Chara = ""
Dim Num As Integer
Dim ColNum As Long
ColNum = InputBox("Input the column number")
If ColNum < 27 Then
Chara = Chr(ColNum + 64) & Chara
Num = ColNum / 26
If (Num * 26) > ColNum Then Num = Num - 1
If (Num * 26) = ColNum Then Num = ((ColNum - 1) / 26) - 1
Chara = Chr((ColNum - (26 * Num)) + 64) & Chara
ColNum = Num
MsgBox "Address is '" & Chara & "'."
LATEST UPDATE: Please ignore the function below, @SurasinTancharoen managed to alert me that it is broken at n = 53.
For those who are interested, here are other broken values just below n = 200:
n = 53
n = 200
Please use @brettdj function for all your needs. It even works for Microsoft Excel latest maximum number of columns limit: 16384 should gives XFD
END OF UPDATE
The function below is provided by Microsoft:
Function ConvertToLetter(iCol As Integer) As String
Dim iAlpha As Integer
Dim iRemainder As Integer
iAlpha = Int(iCol / 27)
iRemainder = iCol - (iAlpha * 26)
If iAlpha > 0 Then
ConvertToLetter = Chr(iAlpha + 64)
If iRemainder > 0 Then
ConvertToLetter = ConvertToLetter & Chr(iRemainder + 64)
Source: How to convert Excel column numbers into alphabetical characters
For reference, this pukes with larger column sets as Chr() doesn't handle large numbers well.
@Azuvector will this work for values less than 100?
This has a bug. Try ConvertToLetter(53) which should have been 'BA' but it will be fail.
@SurasinTancharoen Thank you very much for noting this flaw. I have never thought Microsoft would provide a broken function as they are the one who created Microsoft Excel themselves. I will abandon this function from now on and will use @brettdj function that even correct up to latest Microsoft Excel maximum number of column limit Col_Letter(16384) = "XFD"
Col_Letter(16384) = "XFD"
And where on Earth does this "divide by 27" comes from? Last I checked there are 26 letters. This is why this code breaks.
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
results ALL asuming F15 contains result of step 1
In one go we can write
The solution from brettdj works fantastically, but if you are coming across this as a potential solution for the same reason I was, I thought that I would offer my alternative solution.
The problem I was having was scrolling to a specific column based on the output of a MATCH() function. Instead of converting the column number to its column letter parallel, I chose to temporarily toggle the reference style from A1 to R1C1. This way I could just scroll to the column number without having to muck with a VBA function. To easily toggle between the two reference styles, you can use this VBA code:
If Application.ReferenceStyle = xlR1C1 Then
Application.ReferenceStyle = xlA1
Application.ReferenceStyle = xlR1C1
This is available through using a formula:
and so also can be written as a VBA function as requested:
Function ColName(colNum As Integer) As String
ColName = Split(Worksheets(1).Cells(1, colNum).Address, "$")(1)
Easy way to get the column name
column = Replace(cell.Address(False, False), cell.Row, "")
I hope it helps =)
This will work regardless of what column inside your one code line for cell thats located in row X, in column Y:
Mid(Cells(X,Y).Address, 2, instr(2,Cells(X,Y).Address,"$")-2)
If you have a cell with unique defined name "Cellname":
Mid(Cells(1,val(range("Cellname").Column)).Address, 2, instr(2,Cells(1,val(range("Cellname").Column)).Address,"$")-2)
Here is a simple one liner that can be used.
ColumnLetter = Mid(Cells(Row, LastColA).Address, 2, 1)
It will only work for a 1 letter column designation, but it is nice for simple cases. If you need it to work for exclusively 2 letter designations, then you could use the following:
ColumnLetter = Mid(Cells(Row, LastColA).Address, 2, 2)
If you'd rather not use a range object:
Function ColumnLetter(ColumnNumber As Long) As String
Dim n As Long
Dim c As Byte
Dim s As String
n = ColumnNumber
c = ((n - 1) Mod 26)
s = Chr(c + 65) & s
n = (n - c) \ 26
Loop While n > 0
ColumnLetter = s
Not clear why you posted a longer method with a loop on the basis of If you'd rather not use a range object:
@brettdj I can imagine several reasons: 1) this method is around 6x faster by my testing 2) it doesn't require access to the Excel API 3) it presumably has a smaller memory footprint. EDIT: Also, I'm not sure why I commented on an answer over a year old :S
@blackhawk, fair point re the speed. -1 removed.
There's a drawback to the increased speed, though. Using the range object throws an error if you pass in an invalid column number. It works even if someone is still using Excel 2003. If you need that kind of exception, go with the range method. Otherwise, kudos to robartsd.
You could always test the input column number: IF ColumnNumber <= 16384 Then
IF ColumnNumber <= 16384 Then
IF ColumnNumber <= Columns.Count would be better to avoid assumptions around versions.
IF ColumnNumber <= Columns.Count
For VBS users: If you get an error Expected ')', you will need to remove the As <Type> statements. (As Long, As Byte, As String, etc)
Another reason to use this code is if you're not in VBA but in VB, .net, etc.
Here's another way:
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)
No need to create a string listing the letters of the alphabet. ASCII have essentially done that for us.
There is a very simple way using Excel power: Use Range.Cells.Address property, this way:
strCol = Cells(1, lngRow).Address(xlRowRelative, xlColRelative)
This will return the address of the desired column on row 1. Take it of the 1:
strCol = Left(strCol, len(strCol) - 1)
Note that it so fast and powerful that you can return column addresses that even exists!
Substitute lngRow for the desired column number using Selection.Column property!
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)
lRemainder = iCol Mod 26
lAlpha = Int(iCol / 26)
If lRemainder = 0 Then
lRemainder = 26
lAlpha = lAlpha - 1
ColumnNumberToLetter = ColumnNumberToLetter(lAlpha) & Chr(lRemainder + 64)
Cut-and-paste perfect to convert numbers greater than 676. Thanks!
The remainder can never be more than 26 so why not an integer rather than long?
@Caltor Unless you have a special purpose for using an Integer, like calling an API that demands one for example, you should never choose an Integer over a Long. VBA is optimized for Longs. VBA processes Longs faster than Integers.
@ExcelHero I didn't know that. Doesn't a Long take more memory than an Integer though?
@Caltor Indeed a Long is 32 bits, while an Integer is 16. But that does not matter in modern computing. 25 years ago... it mattered a lot. But today (even 15 years ago) the difference is totally inconsequential.
Something that works for me is:
This will return the $AE$1 format reference for you.