OpenFileDialog.OpenFile Method ()
.NET Framework (current version)
Opens the file selected by the user, with read-only permission. The file is specified by the FileName property.
Assembly: System.Windows.Forms (in System.Windows.Forms.dll)
| Exception | Condition |
|---|---|
| ArgumentNullException | The file name is null. |
The OpenFile method is used to provide a facility to quickly open a file from the dialog box. The file is opened in read-only mode for security purposes. To open a file in read/write mode, you must use another method, such as FileStream.
The following code example demonstrates how to use the OpenFile method.
private void button1_Click(object sender, System.EventArgs e) { Stream myStream = null; OpenFileDialog openFileDialog1 = new OpenFileDialog(); openFileDialog1.InitialDirectory = "c:\\" ; openFileDialog1.Filter = "txt files (*.txt)|*.txt|All files (*.*)|*.*" ; openFileDialog1.FilterIndex = 2 ; openFileDialog1.RestoreDirectory = true ; if(openFileDialog1.ShowDialog() == DialogResult.OK) { try { if ((myStream = openFileDialog1.OpenFile()) != null) { using (myStream) { // Insert code to read the stream here. } } } catch (Exception ex) { MessageBox.Show("Error: Could not read file from disk. Original error: " + ex.Message); } } }
FileDialogPermission
to open a file. Associated enumeration: FileDialogPermissionAccess.Open.
.NET Framework
Available since 1.1
Available since 1.1
Show: