R
robert d via AccessMonster.com
I have an AutoExec macro that opens my StartAppForm upon loading of my
application. I generally develop my app in Access 2002 (default file format
Access 2000) When I'm on the road, I do development in Access 2000. My
client still has Access 2000.
After developing in Access 2000 for awhile I always seem to reach a point
where I get a failure on my AutoExec. It is an "Action Failed" message that
is shown, with the following:
Macro Name : AutoExec
Condition: True
Action Name: OpenForm
Arguments: StartAppForm, Form,,,,Hidden
I have to click on the "Halt" button to proceed.
If I open up this copy of my app in Access 2002, then the problem usually
goes away. I have never run into this problem with Access 2002 (A2002 has
its own problems, though)
The only thing the AutoExec does is to open the StartAppForm. So I have
tried to double-click on the StartAppForm in the database window, after
placing a code break in the code. When I step through the code, I can't find
an error and if I then hit F5 to continue, the app opens normally??!
So does anyone have any suggestions as to how I can try to locate the problem?
Maybe it is just a form of corruption that will go away, but I don't know
that for sure.
application. I generally develop my app in Access 2002 (default file format
Access 2000) When I'm on the road, I do development in Access 2000. My
client still has Access 2000.
After developing in Access 2000 for awhile I always seem to reach a point
where I get a failure on my AutoExec. It is an "Action Failed" message that
is shown, with the following:
Macro Name : AutoExec
Condition: True
Action Name: OpenForm
Arguments: StartAppForm, Form,,,,Hidden
I have to click on the "Halt" button to proceed.
If I open up this copy of my app in Access 2002, then the problem usually
goes away. I have never run into this problem with Access 2002 (A2002 has
its own problems, though)
The only thing the AutoExec does is to open the StartAppForm. So I have
tried to double-click on the StartAppForm in the database window, after
placing a code break in the code. When I step through the code, I can't find
an error and if I then hit F5 to continue, the app opens normally??!
So does anyone have any suggestions as to how I can try to locate the problem?
Maybe it is just a form of corruption that will go away, but I don't know
that for sure.