I try to open a form in another database by using GetObject. Unfortunately I have to open a second instance of the database, but I would like to use the active instance of that database instead (if loaded). TO accomplish this I need to set an object reference to the running instance of that db.
What I currently use is the function below. This function first tries to activate the running instance of the database using its screen name, and if this generates an error the database and the form are loaded. However, if the database is already loaded I want to be able to load the form as well.
On lesser problem is if the error procedure to load the db and form generates an error, the error routine is not followed. How should I manage that?
Anyone has an idea?
I'm Using Access 2016
Thx.
Peter
Public Function AppDbOpen(strAppExec As String, strAppName As String, strOpenForm As String) As Boolean
On Error GoTo Err_Proc
Dim objDb As Object
'Activate DB if open
AppActivate strAppName
AppDbOpen = True
Exit_Err_Proc:
Set objDb = Nothing
Exit Function
Err_Proc:
Select Case Err.Number
Case 5 'Open Db if not open
Set objDb = GetObject(strAppExec, "Access.Application")
If Nz(strOpenForm, "") <> "" Then
objDb.DoCmd.OpenForm strOpenForm
End If
AppDbOpen = True
Case Else
MsgBox "Error: " & Trim(Str(Err.Number)) & vbCrLf & _
"Desc: " & Err.description & vbCrLf & vbCrLf & _
"Module: Mod_GeneralFunctions" & vbCrLf & _
"Function: AppDbOpen", _
vbCritical, "Error!"
End Select
Resume Exit_Err_Proc
End Function