OpenFileDialog.OpenFile Metode

Definisi

Ruang nama:

System.Windows.Forms

Rakitan:

System.Windows.Forms.dll

Sumber:

OpenFileDialog.cs

Membuka file yang dipilih oleh pengguna, dengan izin baca-saja. File ditentukan oleh FileName properti .

public:
 System::IO::Stream ^ OpenFile();

public System.IO.Stream OpenFile();

member this.OpenFile : unit -> System.IO.Stream

Public Function OpenFile () As Stream

Mengembalikan

Stream

Yang Stream menentukan file baca-saja yang dipilih oleh pengguna.

Pengecualian

ArgumentNullException

Nama file adalah null.

IOException

Terjadi kesalahan I/O saat membuka file.

Contoh

Contoh kode berikut menunjukkan cara menggunakan OpenFile metode .

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

Keterangan

Metode OpenFile ini digunakan untuk menyediakan fasilitas untuk membuka file dengan cepat dari kotak dialog. File dibuka dalam mode baca-saja untuk tujuan keamanan. Untuk membuka file dalam mode baca/tulis, Anda harus menggunakan metode lain, seperti FileStream.

Lihat juga

• Stream
• CheckFileExists
• ReadOnlyChecked