How to Use OpenFileDialog in C#

Submitted by janobe on Wednesday, May 22, 2019 - 20:38.

In this tutorial, I’m going to teach you how to use OpenFileDialog in C#. This kind of method allows you to browse and select files on your computer in a form of a windows dialog box. After choosing a file and selects OK, the file that you have chosen will display its contents in the form’s text box.

Creating Application

Step 1

Open Microsoft Visual Studio 2015 and create a new windows form application for c#. ps1

Step 2

Add a Richtexbox,TextBox, button and a OpenfileDialog into the form. Do the form just like shown below. ps2

Step 3

Press F7 to open the code editor. In the code editor, create a method to open the windows dialog box and select a file to display it inside the richtextbox.

   
        private void Open_File()
        {
               
            //    'CHECK THE SELECTED FILE IF IT EXIST OTHERWISE THE DIALOG BOX WILL DISPLAY A WARNING.
            openFileDialog1.CheckFileExists = true;
 
 
            //    'CHECK THE SELECTED PATH IF IT EXIST OTHERWISE THE DIALOG BOX WILL DISPLAY A WARNING.
            openFileDialog1.CheckPathExists = true;
 
 
            //    'GET AND SET THE DEFAULT EXTENSION
            openFileDialog1.DefaultExt = "txt";
 
 
            //    'RETURN THE FILE LINKED TO THE LNK FILE
            openFileDialog1.DereferenceLinks = true;
 
            //    'SET THE FILE NAME TO EMPTY 
            openFileDialog1.FileName = "";
 
            //    'FILTERING THE FILES
            openFileDialog1.Filter = "Text files (*.txt)|*.txt|All files|*.*";
 
            //    'SET THIS FOR ONE FILE SELECTION ONLY.
            openFileDialog1.Multiselect = false;
 
 
 
            //    'SET THIS TO PUT THE CURRENT FOLDER BACK TO WHERE IT HAS STARTED.
            openFileDialog1.RestoreDirectory = true;
 
            //    'SET THE TITLE OF THE DIALOG BOX.
            openFileDialog1.Title = "Select a file to open";
 
            //    'ACCEPT ONLY THE VALID WIN32 FILE NAMES.
            openFileDialog1.ValidateNames = true;
 
            if (openFileDialog1.ShowDialog() == DialogResult.OK)
            {
                richTextBox1.Text = System.IO.File.ReadAllText(openFileDialog1.FileName);
                textBox1.Text = openFileDialog1.FileName;
            } 
        }

Step 4

Write the following code to execute the method that you have created when the button is clicked.

    
        private void button1_Click(object sender, EventArgs e)
        {
            Open_File();
        }

For any questions about this article. You can contact me @ Email – [email protected] Mobile No. – 09305235027 – TNT Or feel free to comment below.

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good

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