Did you run the Auto_Open procedure after you made a change to your code?
If no, then run that Auto_Open. Then do some experimentation with your other
procedure.
ps. I hope that this is a learning exercise. You'd really want to use
ThisWorkbook.Name in other procedures.
VBA beginner wrote:
>
> Unfortunately I can't make this work. I have this public variable in separate
> module and still the value of the fname variable is empty in the other
> module.
>
> "Gary Keramidas" kirjoitti:
>
> > oh, and don't use filename, use fname or some other variant. filename is valid
> > property name in vba.
> >
> > --
> >
> >
> > Gary
> >
> >
> > "Gary Keramidas" <GKeramidasATmsn.com> wrote in message
> > news:(E-Mail Removed)...
> > > try this, but i usually put all of the public variables in a separate module
> > >
> > > Option Explicit
> > > Public filename As String
> > > Sub auto_open()
> > >
> > > filename = Application.ThisWorkbook.Name
> > >
> > > End Sub
> > >
> > >
> > >
> > > --
> > >
> > >
> > > Gary
> > >
> > >
> > > "VBA beginner" <(E-Mail Removed)> wrote in message
> > > news:802D9FE7-F54F-4EDD-AE75-(E-Mail Removed)...
> > >>I have tried to save the name of the opened file to variable, so that I can
> > >> use it in other modules and worksheets. I have the following auto_open sub:
> > >>
> > >>
> > >> Sub auto_open()
> > >> Global filename As Variant
> > >> filename = Application.ThisWorkbook.Name
> > >>
> > >> End Sub
> > >>
> > >> This doesn't work, I have the Compile error message: Invalid attribute in
> > >> Sub or Function. Auto_open sub in the module, so what is wrong in this code?
> > >> Or is there any other way to do this?
> > >>
> > >>
> > >
> > >
> >
> >
> >
--
Dave Peterson
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