Need VBA code for "Open Dialogue" box

[Guest]

Good morning.

I am using an Access VBA module to control Excel and I need to figure out
code that can simply change current directory to a set directory and then
show Excel's "Open" Diaglogue box. I am aware that I can use an open command
to open a certain file, but I don't want to do that. I want dialogue box and
not specify a file. Any suggestions? Thanks

Cordially,

 

[Guest]

hi,
i think it is:
Application.Dialogs(xlDialogOpen).Show

 

[Mark]

Hi,

chDir "c:\Here\Mummy"

application.dialogs("Open").show
workbooks.open Filename

Filename is the filename

Something like this. You have to specify the filename
property then

If you are game, you can steal my code from one of my
utilities: EasyText_Rev1.xls

at:
http://www.geocities.com/excelmarksway

The code is not password protected but you will have to
modify it to suit your needs.


- mark