OpenFileDialog.OpenFile Method

  • Reference

Definition

Namespace:
System.Windows.Forms
Assembly:
System.Windows.Forms.dll

Opens the file selected by the user, with read-only permission. 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

Returns

Stream

A Stream that specifies the read-only file selected by the user.

Exceptions

ArgumentNullException

The file name is null.

IOException

An I/O error occurred while opening the file.

Examples

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

Remarks

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.

Applies to

See also