OpenFileDialog.OpenFile OpenFileDialog.OpenFile OpenFileDialog.OpenFile OpenFileDialog.OpenFile Method

Definición

Abre el archivo seleccionado por el usuario con permiso de solo lectura.Opens the file selected by the user, with read-only permission. La propiedad FileName especifica el archivo.The file is specified by the FileName property.

public:
 System::IO::Stream ^ OpenFile();
public System.IO.Stream OpenFile ();
member this.OpenFile : unit -> System.IO.Stream
Public Function OpenFile () As Stream

Devoluciones

Stream que especifica el archivo de solo lectura seleccionado por el usuario.A Stream that specifies the read-only file selected by the user.

Excepciones

El nombre de archivo es null.The file name is null.

Ejemplos

En el ejemplo de código siguiente se muestra cómo utilizar el método OpenFile.The following code example demonstrates how to use the OpenFile method.

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

Comentarios

El OpenFile método se usa para facilitar una forma de abrir rápidamente un archivo desde el cuadro de diálogo.The OpenFile method is used to provide a facility to quickly open a file from the dialog box. El archivo se abre en modo de solo lectura por motivos de seguridad.The file is opened in read-only mode for security purposes. Para abrir un archivo en modo de lectura/escritura, debe usar otro método, como FileStream.To open a file in read/write mode, you must use another method, such as FileStream.

Seguridad

FileDialogPermission
Para abrir un archivo.to open a file. Enumeración asociada: Open.Associated enumeration: Open.

Se aplica a

Consulte también: