FileDialog.RestoreDirectory 属性

定义

命名空间:

System.Windows.Forms

程序集:

System.Windows.Forms.dll

获取或设置一个值，该值指示该对话框在关闭前是否将目录还原为之前选定的目录。

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

属性值

Boolean

假如用户在搜索文件的过程中更改了目录，那么，如果对话框将当前目录还原为之前选定的目录，则为 true；否则为 false。 默认值是 false。

示例

下面的代码示例使用 OpenFileDialog 实现 FileDialog 并演示如何创建、设置属性以及显示对话框。 该示例使用 RestoreDirectory 该属性来确保对话框关闭时还原以前选择的目录。 本示例需要一个窗体，其中放置了一个 Button 窗体，以及 System.IO 添加到它的命名空间。

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

另请参阅

• InitialDirectory