Tried this code on my own program. Got the same thing. Did some debugging and found that if it successfuly opens the file then 'r' will return '33' not '0', if it unsuccessfully opens the file, then 'r' will return '2'. The following code works flawlessly for me: Dim r As String 'Variable to store the response string Msg = "Failed to open file! " Style = vbOKOnly + vbCritical + vbDefaultButton1 + vbApplicationModal Title = "ERROR!" r = OpenThisFile("c:\test.txt") If r <> 33 Then Response = MsgBox(Msg, Style, Title) End If Hope that helps. ~Scott
