[C#.net] Open File via Dialog / print on textbox

[Chad]

To lazy to make a tutorial

-what you need
using System.IO; (top of code)
OpenFileDialog (renamed dialog)
TextBox (txtData)

  txtData.Text = "";
 
            string strFileLoc = "";
 
            dialog.Filter = "txt files (*.txt)|*.txt|All files (*.*)|*.*";
            dialog.InitialDirectory = "C:"; dialog.Title = "Open";
 
            if (dialog.ShowDialog() == DialogResult.OK)
            {
                strFileLoc = dialog.FileName;
 
                try
                {
                   using (StreamReader sr = new StreamReader(strFileLoc))
                    {
                        string line;
                        while ((line = sr.ReadLine()) != null)
                        {
                            txtData.Text += line + "\r\n";
                        }
                    }
                }
                catch (Exception ex)
                {
                    MessageBox.Show("Failed to open document. Maybe it doesn't exist or the data is corrupted?", "Error");
                }
 
           }
           else if (strFileLoc == String.Empty)
           {
                 return;
           } 
  

Uses FileDialog to find the .txt file the user wants to open, saves the location in a string, StreamReader opens / reads all info in the .txt file.