Opening a file within a file using VB?

[Gabbelgak]

Good evening, I'm a bit of a novice with access but I'm trying to work on automating some mundane tasks at work.

I want to make a form in access that I can press a button it will switch to my work program, navigate to the correct part (I have this working with AppActivate and SendKeys) but I need to open up files within the program and am not sure how to navigate to the correct file and open it with my limited knowledge. I can get to an open dialog box, and can even get that working fine but I don't want to use the default folder it opens to - how would I make it open something located in say C:\Windows\Test\openthis.ab?

I've attached my button code so you can see what I'm thinking here. I'm sure there are probably better ways to go about this but I'm not very experienced

Private Sub Command0_Click()
AppActivate ("AppNameHere")
SendKeys "%N"
SendKeys "R"
SendKeys "%D"
SendKeys "E"
Dim T1 As Variant
Dim T2 As Variant
 T1 = Now()
 T2 = DateAdd("s", 2, T1)
 Do Until T2 <= T1
 T1 = Now()
Loop
SendKeys "%F"
SendKeys "O" 'A windows "Open" dialog box opens here
 T1 = Now()
 T2 = DateAdd("s", 1, T1)
 Do Until T2 <= T1
 T1 = Now()
Loop
'here I can do SendKeys again to make it type something and open something in that folder but I want to open something elsewhere.
End Sub

 

[Micron]

SendKeys is thought by many to be unreliable, thus I've rarely used it. Why don't you just open the file from Access and let Windows handle the decision on which application needs to open it? That would also work around opening to a default folder, no? You might want to look into the msoFileDialogFolderPicker or msoFileDialogFilePicker to see if you can make use of either. You can probably guess which one does what by the name.