Question

Using C#, I need help understanding how to use the OpenFileDialog Box to load a file to my program.

Answer 1

Dear Student ,

As per requirement submitted above kindly find below solution.

Here new Windows Forms Application in C# is created using Visual Studio 2019 with name "OpenFileApplication".This application contains a form with name "Form1.cs".Below is the details of this form.

Form1.cs

//namespace
using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Windows.Forms;
using System.IO;
//application namespace
namespace OpenFileApplication
{
public partial class Form1 : Form
{
public Form1()
{
InitializeComponent();
}
//Form load event
private void Form1_Load(object sender, EventArgs e)
{
//creating object of OpenFileDialog
OpenFileDialog openFileDialog = new OpenFileDialog();
//this line is used to show dialog
if(openFileDialog.ShowDialog()==DialogResult.OK)
{
//reading file contents using StreamReader
StreamReader reader = new StreamReader(openFileDialog.FileName);
//display file contents in the textbox
textBox1.Text = reader.ReadToEnd();
}
}
}
}

==================================

Output :Run application using F5 and will get the screen as shown below

Screen 1:example.txt

Screen 2:

NOTE :PLEASE FEEL FREE TO PROVIDE FEEDBACK ABOUT THE SOLUTION.