How to call “Open file” form

// add system
using System.IO

private void button4_Click(object sender, EventArgs e)
 {
        OpenFileDialog ofd = new OpenFileDialog();   // open file dialog form
         ofd.Filter = "XML|*.xml";                   // filter file can be read

        if (ofd.ShowDialog() == DialogResult.OK)     // if everything ok, run the process
            {
               // insert the value you need from the file
            }
 }

1. Open file form look like