FileDialog.RestoreDirectory Property
Definition
Important
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Gets or sets a value indicating whether the dialog box restores the directory to the previously selected directory before closing.
public:
property bool RestoreDirectory { bool get(); void set(bool value); };
public bool RestoreDirectory { get; set; }
member this.RestoreDirectory : bool with get, set
Public Property RestoreDirectory As Boolean
Property Value
true
if the dialog box restores the current directory to the previously selected directory if the user changed the directory while searching for files; otherwise, false
. The default value is false
.
Examples
The following code example uses the OpenFileDialog implementation of FileDialog and illustrates creating, setting of properties, and showing the dialog box. The example uses the RestoreDirectory property to ensure that the previously selected directory is restored when the dialog box is closed. The example requires a form with a Button placed on it and the System.IO namespace added to it.
private:
void button1_Click( Object^ /*sender*/, System::EventArgs^ /*e*/ )
{
Stream^ myStream;
OpenFileDialog^ openFileDialog1 = gcnew OpenFileDialog;
openFileDialog1->InitialDirectory = "c:\\";
openFileDialog1->Filter = "txt files (*.txt)|*.txt|All files (*.*)|*.*";
openFileDialog1->FilterIndex = 2;
openFileDialog1->RestoreDirectory = true;
if ( openFileDialog1->ShowDialog() == System::Windows::Forms::DialogResult::OK )
{
if ( (myStream = openFileDialog1->OpenFile()) != nullptr )
{
// Insert code to read the stream here.
myStream->Close();
}
}
}
var fileContent = string.Empty;
var filePath = string.Empty;
using (OpenFileDialog openFileDialog = new OpenFileDialog())
{
openFileDialog.InitialDirectory = "c:\\";
openFileDialog.Filter = "txt files (*.txt)|*.txt|All files (*.*)|*.*";
openFileDialog.FilterIndex = 2;
openFileDialog.RestoreDirectory = true;
if (openFileDialog.ShowDialog() == DialogResult.OK)
{
//Get the path of specified file
filePath = openFileDialog.FileName;
//Read the contents of the file into a stream
var fileStream = openFileDialog.OpenFile();
using (StreamReader reader = new StreamReader(fileStream))
{
fileContent = reader.ReadToEnd();
}
}
}
MessageBox.Show(fileContent, "File Content at path: " + filePath, MessageBoxButtons.OK);
Private Sub button1_Click(ByVal sender As Object, ByVal e As System.EventArgs)
Dim myStream As Stream = Nothing
Dim openFileDialog1 As New OpenFileDialog()
openFileDialog1.InitialDirectory = "c:\"
openFileDialog1.Filter = "txt files (*.txt)|*.txt|All files (*.*)|*.*"
openFileDialog1.FilterIndex = 2
openFileDialog1.RestoreDirectory = True
If openFileDialog1.ShowDialog() = System.Windows.Forms.DialogResult.OK Then
Try
myStream = openFileDialog1.OpenFile()
If (myStream IsNot Nothing) Then
' Insert code to read the stream here.
End If
Catch Ex As Exception
MessageBox.Show("Cannot read file from disk. Original error: " & Ex.Message)
Finally
' Check this again, since we need to make sure we didn't throw an exception on open.
If (myStream IsNot Nothing) Then
myStream.Close()
End If
End Try
End If
End Sub