How to Open File Dialog in C#

The OpenFileDialog object interacts with the Computer’s API (Application Programming Interface) to present available files to the user and retrieves the user’s file selection back to the program. This object is part of the System.Windows.Forms library so no additional using reference will be needed.

This example uses a button click method.

private void btnOpenTextFile_Click(object sender, EventArgs e)

{

//First, declare a variable to hold the user’s file selection.

private void String input = string.Empty;

//Create a new instance of the OpenFileDialog because it's an object.

private voidOpenFileDialog dialog = new OpenFileDialog();

//Now set the file type

dialog.Filter = "txt files (*.txt)|*.txt|All files (*.*)|*.*";

//Set the starting directory and the title.

dialog.InitialDirectory = "C:"; dialog.Title = "Select a text file";

//Present to the user.

if (dialog.ShowDialog() == DialogResult.OK)

     strFileName = dialog.FileName;

if (strFileName == String.Empty)

     return;//user didn't select a file

}