Opening multiple files (OpenFileDialog, C#)

Question

I'm trying to open multiple files at once with the OpenFileDialog, using FileNames instead of FileName. But I cannot see any examples anywhere on how to accomplish this, not even on MSDN. As far as I can tell - there's no documentation on it either. Has anybody done this before?

Answer 1

You must set the OpenFileDialog.Multiselect Property value to true, and then access the OpenFileDialog.FileNames property.

Check this sample

private void Form1_Load(object sender, EventArgs e) {     InitializeOpenFileDialog(); }  private void InitializeOpenFileDialog() {     // Set the file dialog to filter for graphics files.     this.openFileDialog1.Filter =         "Images (*.BMP;*.JPG;*.GIF)|*.BMP;*.JPG;*.GIF|" +         "All files (*.*)|*.*";      //  Allow the user to select multiple images.     this.openFileDialog1.Multiselect = true;     //                   ^  ^  ^  ^  ^  ^  ^      this.openFileDialog1.Title = "My Image Browser"; }  private void selectFilesButton_Click(object sender, EventArgs e) {     DialogResult dr = this.openFileDialog1.ShowDialog();     if (dr == System.Windows.Forms.DialogResult.OK)     {         // Read the files         foreach (String file in openFileDialog1.FileNames)          {             // Create a PictureBox.             try             {                 PictureBox pb = new PictureBox();                 Image loadedImage = Image.FromFile(file);                 pb.Height = loadedImage.Height;                 pb.Width = loadedImage.Width;                 pb.Image = loadedImage;                 flowLayoutPanel1.Controls.Add(pb);             }             catch (SecurityException ex)             {                 // The user lacks appropriate permissions to read files, discover paths, etc.                 MessageBox.Show("Security error. Please contact your administrator for details.\n\n" +                     "Error message: " + ex.Message + "\n\n" +                     "Details (send to Support):\n\n" + ex.StackTrace                 );             }             catch (Exception ex)             {                 // Could not load the image - probably related to Windows file system permissions.                 MessageBox.Show("Cannot display the image: " + file.Substring(file.LastIndexOf('\\'))                     + ". You may not have permission to read the file, or " +                     "it may be corrupt.\n\nReported error: " + ex.Message);             }         }     } 

Answer 2

You can use this method for text files:

OpenFileDialog open = new OpenFileDialog(); open.Filter = "All Files *.txt | *.txt"; open.Multiselect = true; open.Title = "Open Text Files";  if (open.ShowDialog() == DialogResult.OK) {     foreach (String file in open.FileNames)     {              string temp = YourRichTextBox.Text;           YourRichTextBox.LoadFile(file, RichTextBoxStreamType.PlainText);           YourRichTextBox.Text += temp;     } }