carlAlex
8/17/2016 - 12:30 PM

C# OpenFileDialog

C# OpenFileDialog

private void button1_Click(object sender, EventArgs e)
	{
	    // Show the dialog and get result.
	    DialogResult result = openFileDialog1.ShowDialog();
	    if (result == DialogResult.OK) // Test result.
	    {
	    }
	    Console.WriteLine(result); // <-- For debugging use.
	}