OpenFileDialog

On a form create a textbox (txtFilePath), button (btnFilePathBrowse), and OpenFileDialog (ofdFilePath).

Double click on the button and the OpenFileDialog in Design view to create the action scripts

Private Sub btnFilePathBrowse_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles btnFilePathBrowse.Click
		ofdFilePath.Title = "Please Select a File"
		ofdFilePath.FileName = ""
		ofdFilePath.InitialDirectory = "C:"
		ofdFilePath.ShowDialog()
End Sub

Private Sub ofdFilePath_FileOk(ByVal sender As System.Object, ByVal e As System.ComponentModel.CancelEventArgs) Handles ofdFilePath.FileOk
	Dim sioStream As System.IO.Stream
	sioStream = ofdFilePath.OpenFile()
	txtFilePath.Text = ofdFilePath.FileName.ToString()
End Sub

private void btnFilePathBrowse_Click(System.Object sender, System.EventArgs e)
{
	ofdFilePath.Title = "Please Select a File";
	ofdFilePath.FileName = "";
	ofdFilePath.InitialDirectory = "C:";
	ofdFilePath.ShowDialog();
}

private void ofdFilePath_FileOk(System.Object sender, System.ComponentModel.CancelEventArgs e)
{
	System.IO.Stream sioStream = null;
	sioStream = ofdFilePath.OpenFile();
	txtFilePath.Text = ofdFilePath.FileName.ToString();
}