OpenFileDialog
我正在尝试使用 ,FileNames
而不是.一次打开多个文件FileName
。但是我在任何地方都看不到任何关于如何做到这一点的例子,甚至在 MSDN 上也看不到。据我所知 - 也没有关于它的文档。以前有人做过吗?
问问题
77049 次
2 回答
79
您必须将 OpenFileDialog.Multiselect
属性值设置为 true,然后才能访问该OpenFileDialog.FileNames
属性。
检查这个样本
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);
}
}
}
于 2009-08-21T12:13:16.400 回答
1
您可以将此方法用于文本文件:
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;
}
}
于 2020-05-11T10:42:10.143 回答