Sub Open File In VB.NET

    Private Sub OpenFile(ByVal FileName As String)
        Try
            Dim p As New System.Diagnostics.Process
            Dim s As New System.Diagnostics.ProcessStartInfo(FileName)
            s.UseShellExecute = True
            s.WindowStyle = ProcessWindowStyle.Normal
            p.StartInfo = s
            p.Start()
        Catch ex As Exception
            MessageBox.Show("ไม่พบไฟล์'" & FileName & "' ในระบบ!", Me.Text, MessageBoxButtons.OK, MessageBoxIcon.Error)
        End Try
    End Sub