Dim MdbPath As String = "c:\abc.mdb"
 
      Using SW As New IO.StreamWriter(MdbPath & "FTPEIN.BAT")
         SW.WriteLine("@echo off")
         'etc
 
      End Using

      Using objShell As New System.Diagnostics.Process
         objShell.StartInfo.FileName = "notepad.exe"
         objShell.StartInfo.Arguments = "c:\test\test.txt"
         objShell.StartInfo.UseShellExecute = False
         objShell.StartInfo.WindowStyle = ProcessWindowStyle.Normal
         objShell.StartInfo.RedirectStandardOutput = True
         objShell.Start()
         objShell.WaitForExit()
      End Using
'-------------------------------------------------------
      Dim Filename As String = "c:\test\test.xls"
      Try
         Using objShell As New System.Diagnostics.Process
            objShell.StartInfo.FileName = Filename
            objShell.StartInfo.Arguments = Nothing
            objShell.StartInfo.UseShellExecute = True
            objShell.StartInfo.WindowStyle = ProcessWindowStyle.Normal
            objShell.StartInfo.RedirectStandardError = False
            objShell.Start()
            objShell.WaitForExit(1000)
         End Using
      Catch ex As Exception
         MessageBox.Show(ex.Message)
      End Try
'---------------------------------------------------------
''' <summary>
''' 'Öffnen mit' Dialog anzeigen
''' </summary>
''' <param name="Filename">Datei, die geöffnet werden soll</param>
Public Sub ShowFileOpenWithDlg(ByVal Filename As String)
  Dim startInfo As New ProcessStartInfo
  With startInfo
    .FileName = "RUNDLL32.EXE"
    .Arguments = "shell32.dll,OpenAs_RunDLL " & Filename
  End With
  Try
    ' Öffnen mit Dialog starten
    Process.Start(startInfo)
  Catch ex As Exception
    MsgBox("Fehler!" & vbCrLf & ex.Message, MsgBoxStyle.Exclamation)
  End Try
End Sub