Ron,

I may not have made myself clear.

I am getting the dialog box when I open the first workbook, WB01, not when
the second, WB02, is opened.

Doesn't this solution address the situation where WB02 was linked to a third
file?

By the way, everytime I post I get the same "there was a problem with your
post......" message. Is this normal?

Thanks again for your help.

Tim Kredlo


"Ron de Bruin" wrote:

Hi Tim

Use it like this in the Thisworkbook module of WB01

Private Sub Workbook_Open()
Dim WB02 As Workbook
Set WB02 = Workbooks.Open("C:\Data\test1.xls", UpdateLinks:=0)
End Sub


--
Regards Ron de Bruin
http://www.rondebruin.nl


"Tim Kredlo" <Tim wrote in message ...
Sorry for the double posting. In fact, I was surprised either of them
appeared since after hitting 'Post' I was told both times something like
"Sorry. There was a problem with your post. Somebody is being notified."
Thanks for the suggestion, now all I have to do is figure out 'how & where
to do' what you suggested. Am I correct in assuming I need to:
1. create a class module. (I named it "claAppEvents")
2. in claAppEvents 'General' code section have something like "Public
WithEvents XL As Application"
3. create a 'regular' module (I name it "modStartup")
4. in modStartup 'General' code section have something like "Dim App As New
claAppEvents"
5. In my Workbook_Open()procedure for the WB01 include "Set App.XL =
Application"
6. Set WB01 = Workbooks.Open(UpdateLinks:=0)







"Ron de Bruin" wrote:

Possible answer in your other thread

--
Regards Ron de Bruin
http://www.rondebruin.nl


"Tim Kredlo" <Tim wrote in message ...
I have a workbook (WB01) whose Workbook_Open() subprocedure opens a second
workbook (WB02). Because WB01 is linked to WB02, the 'This workbook contains
links to other data sources' dialog box is presented to the user. Is there
code that I can put in the Workbook_Open subprocedure to stop this dialog box
from appearing?