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複数のファイルを開く(OpenFileDialog、C#)

OpenFileDialogの代わりにFileNamesを使用して、FileNameで複数のファイルを一度に開こうとしています。しかし、MSDNでさえ、これを達成する方法についての例はどこにも見当たりません。私が知る限り、ドキュメントもありません。誰もこれをやったことがありますか?

28
jay_t55

OpenFileDialog.Multiselect プロパティ値をtrueに設定し、OpenFileDialog.FileNamesプロパティ。

このサンプルを確認してください

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);
            }
        }
    }
68
RRUZ