FileStream Fs = FD.OpenFile();
is sufficient, but this would be read-only...
private void button1_Click(object sender, RoutedEventArgs e)
{
OpenFileDialog FD = new OpenFileDialog();
FD.Filter = "XAML Files (*.xaml)|*.xaml";
if(FD.ShowDialog() == DialogResult.OK)
{
string fileName = FD.FileName;
if (File.Exists(fileName))
{
FileStream Fs = new FileStream(fileName, FileMode.Open);
}
}
}