Errors to fix to next version:
Instead of using:
[php] OpenFileDialog1.ShowDialog()
If OpenFileDialog1.ShowDialog = Windows.Forms.DialogResult.OK Then
'do something
End If[/php]
You can(must) use:
[php] If OpenFileDialog1.ShowDialog = Windows.Forms.DialogResult.OK Then
'do something
End If[/php]
Because your code opens 2 openfiledialogs.