Hi Pal

GetOpenFilename don't open the file (see the Excel help)
You need to open the file

Workbooks.Open(FName)


Sub test()
Dim FName As Variant
Dim wb As Workbook
FName = Application.GetOpenFilename(filefilter:="Excel Files (*.xls), *.xls")
If FName < False Then
Set wb = Workbooks.Open(FName)
MsgBox "your code"
wb.Close
End If
End Sub


--
Regards Ron de Bruin
(Win XP Pro SP-1 XL2000-2003)
www.rondebruin.nl



"Pal" wrote in message news:Cmo0c.423198$I06.4711213@attbi_s01...


Hi
I have the following code that SHOULD open a file
and copy data from range a12:c100 of that opened file into the calling
worksheet at sheet1 A38.
I get a subscript out of range error trying to activate the newly opened
file.

I passed the file name to the msgbox and found it has the directory path in
it.
Since I am on a network is there a way to remove that path so any user at
different
directories can open the file?

Thanks
Pal


Private Sub CommandButton1_Click()
Filename = Application.GetOpenFilename
Windows(Filename).Activate

Range("A12:C100").Select
Selection.Copy
Windows("results sheet").Activate
Range("A38").Select
Selection.PasteSpecial Paste:=xlValues, Operation:=xlNone, SkipBlanks:=
_
False, Transpose:=False
End Sub