L
Lauro
I need to get a reference to an already open instance of Access if it
is running or open it if not.
I'm using GetObject.
If no other instances of Access are open I have no problem. But I
have no guarantee to point to the rigth one if the user has already
opened one.
I found an half solution with a strange behavior:
Set oAccApp = GetObject(MyFileName, "Access.Application")
oAccApp.Visible = True
works fine (it open a new instance if it is not open or open and
"magically" merge it with the old one if already exists) if I don't
have a StartUp Form.
if I have a StartUp form, I tried to use:
Set oAccApp = GetObject(MyFileName, "Access.Application")
oAccApp.Visible = True
oAccApp.DoCmd.Close acForm, "Welcome Form"
But I get a run time error 2455 .
If I delete
oAccApp.Visible = True
I get multiple instances of my database.
Why?
Any Suggestions?
Thanks Lauro
is running or open it if not.
I'm using GetObject.
If no other instances of Access are open I have no problem. But I
have no guarantee to point to the rigth one if the user has already
opened one.
I found an half solution with a strange behavior:
Set oAccApp = GetObject(MyFileName, "Access.Application")
oAccApp.Visible = True
works fine (it open a new instance if it is not open or open and
"magically" merge it with the old one if already exists) if I don't
have a StartUp Form.
if I have a StartUp form, I tried to use:
Set oAccApp = GetObject(MyFileName, "Access.Application")
oAccApp.Visible = True
oAccApp.DoCmd.Close acForm, "Welcome Form"
But I get a run time error 2455 .
If I delete
oAccApp.Visible = True
I get multiple instances of my database.
Why?
Any Suggestions?
Thanks Lauro