OpenFileDialog.OpenFile 方法

定義

命名空間:

System.Windows.Forms

組件:

System.Windows.Forms.dll

以唯讀許可權開啟用戶選取的檔案。 檔案是由 FileName 屬性所指定。

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

public System.IO.Stream OpenFile ();

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

Public Function OpenFile () As Stream

傳回

Stream

Stream，指定用戶選取的唯讀檔案。

例外狀況

ArgumentNullException

檔案名 null。

IOException

開啟檔案時發生 I/O 錯誤。

範例

下列程式代碼範例示範如何使用 OpenFile 方法。

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

備註

OpenFile 方法可用來提供一個工具，以便從對話框快速開啟檔案。 基於安全性考慮，檔案會以唯讀模式開啟。 若要以讀取/寫入模式開啟檔案，您必須使用其他方法，例如 FileStream。

另請參閱

• Stream
• CheckFileExists
• ReadOnlyChecked