Opening specific file using common dialog

[azkabancells]

How do I open a specific file (an access db) using the common dialog function? I presume it needs to be automatically displayed in the file name of the open file box.

Also I want to be able to either open it at the beginning of the file or create new record by opening at the end of the file. I have created a menu with new and open for this purpose but can't find the code I need.

Thanks for looking

 

[bclt]

c is the CommonDialog

c.FileName = ""
c.ShowOpen

If c.FileName <> "" Then
  MsgBox c.FileName
End If

If you select a file AND press OK in "c.FileName" there will be the full path of the selected file (containing "\the_file"). Now you can open it:

1. Using Shell() method (there will be error if you write Shell(c.Filename), because .mdb is not .exe !)
2. Having added a reference to MS Access type object library, open access (visible or not) and then open the specific database.


-bclt

 

[bclt]

Shell("c:\program files\office ...\access.exe" & " " & c.FileName, 1)

In the "..." I input the path where access.exe is, then a space and then the filename.

The "1" is for access; the window to be focused and also to have its normal size. (equivalent = vbNormalFocus)


-bclt

 

[azkabancells]

Thanks. I specifically need to open a db file called address. On Private Sub mnuNew_Click()I need to open it to eof and on Private Sub mnuOpen_Click()I need to open to bof.

I have so far

f = FreeFile
CMDialog1.Filter = "all files|*.*|text|*.txt"
CMDialog1.FilterIndex = 0
CMDialog1.FileName = "C:\Address"
CMDialog1.ShowOpen
MsgBox "You selected: " + CMDialog1.FileName
TextFileNum = FreeFile
OpenFileName = CMDialog1.FileName
Open OpenFileName For Output As #f

Also are there any good tutorial sites on common dialog and linking db's to access via code (not data control).

Thanks for your help