Return activecell's column header in Alphabet
Hi Rob
Here's a way to do it.
Dim strLastColumn As String
Dim L As Integer
strLastColumn = ActiveCell.Address
L = InStr(2, strLastColumn, "$", 1)
strLastColumn = Mid(strLastColumn, 2, L - 2)
Regards,
Per
"Rob" skrev i meddelelsen
...
Tim,
Is there a way to get Excel to return the column in letter form? I am
trying to extract the alphabetical reference to a cell but I keep getting
the
number. I use code like this:
Dim strLastColumn as String
strLastColumn = ActiveCell.Column
The .Column property always returns the number. So, if I am in column Z
it
returns 26 instead of Z. I need to return Z.
Thanks,
Rob
"Tim Williams" wrote:
dim x as integer, sLetter as string
x=28
sLetter = replace(activesheet.cells(1,x).address(false,false ),"1","")
Tim
"Edmund" wrote in message
...
Pls teach me the correct way to "return activecell's column header in
Alphabet".
In VBA:
Columns(1) represents column A
Columns(2) represents column B
Columns(5) represents column E
Columns(26) represents column Z
When I get VBA to return a variable (an integer) between 1~26, how can
I
convert this back to alphabet (A~Z)?
Currently in my codes, I do like this to display the "Alphabet" of the
column header of activecell:
'Msgbox Left(ActiveCell.Address(False, False, xlA1),1)
Now, I get into trouble when the returned integer exceeds 26.
Thanks a lot.
--
Edmund
(Using Excel XP)
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