open file dialog in c# (2 methods)

you created an app that needs to open an external file manually, for example you created a player in c# and you want to let the user open up files that's why you use openfiledialog which prompt a dialog file and let you choose an specific file.
to do here is the code only for openfiledialog 
i will provide you two methods 1.for text files 2. windows media player 
please add using.sytem.IO namespace first


1.method for text :
  //allow you to open a dilogbox and load the file you desire 
            OpenFileDialog record = new OpenFileDialog();
            record.FileName = "";
            record.Filter = "Text Files|*.txt";
            if (record.ShowDialog() == System.Windows.Forms.DialogResult.OK)
            {
                string file = record.FileName;
              string[]   text = File.ReadAllLines(file);
                 
                 listBox2.Items.AddRange(text);

            }


2.method for windows media  player :


using (FileDialog dlgopenfile = new OpenFileDialog())
            {

                dlgopenfile.Filter = "all files|*.*";
                dlgopenfile.FileName = "";
                if (dlgopenfile.ShowDialog() == System.Windows.Forms.DialogResult.OK)// show dilog baraie in ke betavan panjarei barai entekhabe media enekhb kard in tabe be esme modal function ham niz shenakhte mishavad //
                {
                    try
                    {
                        
                        mediaplayer.URL = dlgopenfile.FileName;
                        label1.Text = dlgopenfile.FileName;
                    }
                    catch (Exception ex)
                    {
                        MessageBox.Show("unable to load the file " + ex.Message);

                    }

                }
            }


                

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