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

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.