使用按钮和ajax删除mysql

Question

I want to delete files in my database instantly when I press a button. Files were deleted but I am obliged to refresh my navigator. Ajax doesn't operate.

HTML/PHP

  echo '<button class="delete_video" id="'.$videoId.'" type="button">Delete</button>';

javaScript

   $(document).ready(function()
    {
        $(".delete_video").click(function()
        {
            var del_id = $(this).attr('id');
            $.ajax({
                type:'POST',
                url:'delete.php',
                data:'delete_id='+del_id,
                success: function(data)
                {
                    //confirmation of deletion
                }
            });
        });
    });

PHP

$id = $_POST['delete_id'];
include('functions.php');
$DB = connexion();
$DB->query('DELETE FROM videos WHERE id = "'.$id.'"');

Answer 1

Add location.reload() to the success function.

That is the syntax to use.

Plus, you're also open to a serious SQL injection.

Consult the following:

• http://php.net/manual/en/security.database.sql-injection.php
• How can I prevent SQL injection in PHP?

Answer 2

$(document).ready(function()
    {
        $(".delete_video").click(function()
        {
            var del_id = $(this).attr('id');
            $.ajax({
                type:'POST',
                url:'delete.php',
                data:'delete_id='+del_id,
                success: function(data)
                {
                    //reload page
                    location.reload();
                }
            });
        });
    });

Answer 3

I am hoping there are rows you are deleting so make sure that you pass an id element to each row as

---

here goes your video

---

and when your AJAX return success just make the element disappear

 <div id="del_id"> here goes your video to delete </div> $(document).ready(function() { $(".delete_video").click(function() { var del_id = $(this).attr('id'); $.ajax({ type:'POST', url:'delete.php', data:'delete_id='+del_id, success: function(data) { jQuery('#del_id').fadeOut('slow'); } }); }); });