x = ActiveCell.Column
Returns 1 for column A

How do I get x = A?

Hmmm

Arturo

Arturo,

Function GetColumn(ByRef lngColumnNum As Long) As String
GetColumn = Application.Substitute(Cells(1, _
lngColumnNum).Address(False, False), "1", vbNullString)
End Function
--
Jim Cone
San Francisco, USA
http://www.realezsites.com/bus/primitivesoftware


"Arturo" <(E-Mail Removed)>
wrote in message
x = ActiveCell.Column
Returns 1 for column A
How do I get x = A?
Hmmm
Arturo

One way:

Option Explicit
Function ColLetter(rng As Range) As String
Dim myStr As String
With rng.Parent
myStr = .Cells(1, rng.Column).Address(0, 0)
myStr = Left(myStr, Len(myStr) - 1)
End With
ColLetter = myStr
End Function

And I could test it with:
Sub testme()
MsgBox ColLetter(ActiveCell)
End Sub



Arturo wrote:
>
> x = ActiveCell.Column
> Returns 1 for column A
>
> How do I get x = A?
>
> Hmmm
>
> Arturo

--

Dave Peterson

Thank you.

Dim arr As Variant
arr = Split(ActiveCell.Address, "$")
MsgBox arr(1)

"Dave Peterson" wrote:

> One way:
>
> Option Explicit
> Function ColLetter(rng As Range) As String
> Dim myStr As String
> With rng.Parent
> myStr = .Cells(1, rng.Column).Address(0, 0)
> myStr = Left(myStr, Len(myStr) - 1)
> End With
> ColLetter = myStr
> End Function
>
> And I could test it with:
> Sub testme()
> MsgBox ColLetter(ActiveCell)
> End Sub
>
>
>
> Arturo wrote:
> >
> > x = ActiveCell.Column
> > Returns 1 for column A
> >
> > How do I get x = A?
> >
> > Hmmm
> >
> > Arturo
>
> --
>
> Dave Peterson
>

thanks.

Dim arr As Variant
arr = Split(ActiveCell.Address, "$")
MsgBox arr(1)


"Jim Cone" wrote:

> Arturo,
>
> Function GetColumn(ByRef lngColumnNum As Long) As String
> GetColumn = Application.Substitute(Cells(1, _
> lngColumnNum).Address(False, False), "1", vbNullString)
> End Function
> --
> Jim Cone
> San Francisco, USA
> http://www.realezsites.com/bus/primitivesoftware
>
>
> "Arturo" <(E-Mail Removed)>
> wrote in message
> x = ActiveCell.Column
> Returns 1 for column A
> How do I get x = A?
> Hmmm
> Arturo
>

That won't work in XL 97.
--
Jim Cone
San Francisco, USA
http://www.realezsites.com/bus/primitivesoftware



"Arturo" <(E-Mail Removed)>
wrote in message
thanks.
Dim arr As Variant
arr = Split(ActiveCell.Address, "$")
MsgBox arr(1)



"Jim Cone" wrote:
> Arturo,
> Function GetColumn(ByRef lngColumnNum As Long) As String
> GetColumn = Application.Substitute(Cells(1, _
> lngColumnNum).Address(False, False), "1", vbNullString)
> End Function
--
> Jim Cone
> San Francisco, USA
> http://www.realezsites.com/bus/primitivesoftware



> "Arturo" <(E-Mail Removed)>
> wrote in message
> x = ActiveCell.Column
> Returns 1 for column A
> How do I get x = A?
> Hmmm
> Arturo
>

Was this a test?

Arturo wrote:
>
> Thank you.
>
> Dim arr As Variant
> arr = Split(ActiveCell.Address, "$")
> MsgBox arr(1)
>
> "Dave Peterson" wrote:
>
> > One way:
> >
> > Option Explicit
> > Function ColLetter(rng As Range) As String
> > Dim myStr As String
> > With rng.Parent
> > myStr = .Cells(1, rng.Column).Address(0, 0)
> > myStr = Left(myStr, Len(myStr) - 1)
> > End With
> > ColLetter = myStr
> > End Function
> >
> > And I could test it with:
> > Sub testme()
> > MsgBox ColLetter(ActiveCell)
> > End Sub
> >
> >
> >
> > Arturo wrote:
> > >
> > > x = ActiveCell.Column
> > > Returns 1 for column A
> > >
> > > How do I get x = A?
> > >
> > > Hmmm
> > >
> > > Arturo
> >
> > --
> >
> > Dave Peterson
> >

--

Dave Peterson

That would be the slowest way to do it. Enjoy.

--
Regards,
Tom Ogilvy


"Arturo" <(E-Mail Removed)> wrote in message
news:CB3BCEB1-2B27-4BC2-AEC9-(E-Mail Removed)...
> Thank you.
>
> Dim arr As Variant
> arr = Split(ActiveCell.Address, "$")
> MsgBox arr(1)
>
> "Dave Peterson" wrote:
>
>> One way:
>>
>> Option Explicit
>> Function ColLetter(rng As Range) As String
>> Dim myStr As String
>> With rng.Parent
>> myStr = .Cells(1, rng.Column).Address(0, 0)
>> myStr = Left(myStr, Len(myStr) - 1)
>> End With
>> ColLetter = myStr
>> End Function
>>
>> And I could test it with:
>> Sub testme()
>> MsgBox ColLetter(ActiveCell)
>> End Sub
>>
>>
>>
>> Arturo wrote:
>> >
>> > x = ActiveCell.Column
>> > Returns 1 for column A
>> >
>> > How do I get x = A?
>> >
>> > Hmmm
>> >
>> > Arturo
>>
>> --
>>
>> Dave Peterson
>>