Sounds like you are talking about a column label under A1 style addressing.
If so, you can just use
? Left(Cells(1,36).Address(0,0),1 - (36>26))
AJ
where 36 could be replaced with a variable that holds the column number.

Signature
Regards,
Tom Ogilvy
> something like:
> Function Number2Row(Number As Long) As String
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>
> http://www.microsoft.com/office/community/en-us/default.mspx?mid=9bac9c6a-da06-4
7b3-8298-ae17b295265f&dg=microsoft.public.excel.programming
Markus - 22 May 2006 16:35 GMT
so that instead of a full function will do it jobs
Thank you!
Left(Cells(1, ActiveCell.Column).Address(0, 0), 1 - (ActiveCell.Column > 26))
> Sounds like you are talking about a column label under A1 style addressing.
> If so, you can just use
[quoted text clipped - 48 lines]
> >
> > http://www.microsoft.com/office/community/en-us/default.mspx?mid=9bac9c6a-da06-4
7b3-8298-ae17b295265f&dg=microsoft.public.excel.programming
Markus - 22 May 2006 16:46 GMT
would it be possible to add that as an example in the "help"-file for future
versions?
> Sounds like you are talking about a column label under A1 style addressing.
> If so, you can just use
[quoted text clipped - 48 lines]
> >
> > http://www.microsoft.com/office/community/en-us/default.mspx?mid=9bac9c6a-da06-4
7b3-8298-ae17b295265f&dg=microsoft.public.excel.programming