How to: Display an OpenFileDialog Dynamically

You can use the OpenFileDialog component to enable users to browse to a text file and load the selected file to a RichTextBox control on a Windows Form. This example instantiates OpenFileDialog at run-time.

Example

// Create an OpenFileDialog object.
OpenFileDialog openFile1 = new OpenFileDialog();

// Initialize the OpenFileDialog to look for text files.
openFile1.Filter = "Text Files|*.txt";

// Check if the user selected a file from the OpenFileDialog.
if(openFile1.ShowDialog() == System.Windows.Forms.DialogResult.OK)

    // Load the contents of the file into a RichTextBox control.
    richTextBox1.LoadFile(openFile1.FileName,
    RichTextBoxStreamType.PlainText);

Compiling the Code

• Copy the code into the Load_Form1 event handler. When you run the program, you will be prompted to select a text file. The contents of the selected file will be displayed in a RichTextBox control.

Robust Programming

Use the CheckFileExists, CheckPathExists, DefaultExt, Filter, Multiselect, and ValidateNames properties of the OpenFileDialog control to limit run-time errors.

See Also

Concepts

Designing a User Interface in Visual C#

Other Resources

Dialog Boxes (Visual C#)

Visual C# Guided Tour