docs-desktop/dotnet-desktop-guide/framework/winforms/controls/how-to-load-files-into-the-windows-forms-richtextbox-control.md

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How to: Load Files into the Windows Forms RichTextBox Control

The Windows Forms xref:System.Windows.Forms.RichTextBox control can display a plain-text, Unicode plain-text, or Rich-Text-Format (RTF) file. To do so, call the xref:System.Windows.Forms.RichTextBox.LoadFile%2A method. You can also use the xref:System.Windows.Forms.RichTextBox.LoadFile%2A method to load data from a stream. For more information, see xref:System.Windows.Forms.RichTextBox.LoadFile%28System.IO.Stream%2CSystem.Windows.Forms.RichTextBoxStreamType%29.

To load a file into the RichTextBox control

1. Determine the path of the file to be opened using the xref:System.Windows.Forms.OpenFileDialog component. For an overview, see OpenFileDialog Component Overview.

2. Call the xref:System.Windows.Forms.RichTextBox.LoadFile%2A method of the xref:System.Windows.Forms.RichTextBox control, specifying the file to load and optionally a file type. In the example below, the file to load is taken from the xref:System.Windows.Forms.OpenFileDialog component's xref:System.Windows.Forms.FileDialog.FileName%2A property. If you call the method with a file name as its only argument, the file type will be assumed to be RTF. To specify another file type, call the method with a value of the xref:System.Windows.Forms.RichTextBoxStreamType enumeration as its second argument.

   In the example below, the xref:System.Windows.Forms.OpenFileDialog component is shown when a button is clicked. The file selected is then opened and displayed in the xref:System.Windows.Forms.RichTextBox control. This example assumes a form has a button,btnOpenFile.

   Private Sub btnOpenFile_Click(ByVal sender As System.Object, _
      ByVal e As System.EventArgs) Handles btnOpenFile.Click
        If OpenFileDialog1.ShowDialog() = DialogResult.OK Then
          RichTextBox1.LoadFile(OpenFileDialog1.FileName, _
             RichTextBoxStreamType.RichText)
         End If
   End Sub
   

   private void btnOpenFile_Click(object sender, System.EventArgs e)
   {
      if(openFileDialog1.ShowDialog() == DialogResult.OK)
      {
        richTextBox1.LoadFile(openFileDialog1.FileName, RichTextBoxStreamType.RichText);
      }
   }
   

   private:
      void btnOpenFile_Click(System::Object ^  sender,
         System::EventArgs ^  e)
      {
         if(openFileDialog1->ShowDialog() == DialogResult::OK)
         {
            richTextBox1->LoadFile(openFileDialog1->FileName,
               RichTextBoxStreamType::RichText);
         }
      }
   

   (Visual C#, Visual C++) Place the following code in the form's constructor to register the event handler.

   this.btnOpenFile.Click += new System.EventHandler(this. btnOpenFile_Click);
   

   this->btnOpenFile->Click += gcnew
      System::EventHandler(this, &Form1::btnOpenFile_Click);
   

   Important

   To run this process, your assembly may require a privilege level granted by the xref:System.Security.Permissions.FileIOPermission?displayProperty=nameWithType class. If you are running in a partial-trust context, the process might throw an exception because of insufficient privileges. For more information, see Code Access Security Basics.

See also

• xref:System.Windows.Forms.RichTextBox.LoadFile%2A?displayProperty=nameWithType
• xref:System.Windows.Forms.RichTextBox
• RichTextBox Control
• Controls to Use on Windows Forms