title: submit button on an ajax popup does not work
body: <p>i have an ajax popup like this...</p>
  
  <pre><code>&lt;div class="white-popup-block" style="max-width:600px; margin: 20px auto;"&gt;
  &lt;?php
  include "konekdb.php";
  if(isset($_GET['kode_kb'])) {
      $kode_kb = mysql_real_escape_string($_GET['kode_kb']);
  
          $sql    = "SELECT * FROM  kb_tiny 
                           WHERE kode_kb='$kode_kb'" ;
          $query = mysql_query($sql, $koneksiDB) or die ("error query".mysql_error());
          $data = mysql_fetch_assoc($query);
          echo "Title : ".$data['title'];
          echo "No Ticket :".$data['no_ticket'];
          echo "Error Code : ".$data['error_code'];
          echo "example :  ".$data['ex_case'];
          echo "Solution :".$data['solution'];
          echo "Creator :  ".$data['creator'];
          echo "Date Modified :  ".$data['modified'];
  
  }
  ?&gt;
  &lt;/br&gt;
  &lt;/br&gt;
  &lt;/br&gt;
  &lt;form method="POST" action="&lt;?php echo $_SERVER['PHP_SELF']; ?&gt;"&gt;
  &lt;input name="approved" type="button" value="Approved" /&gt; &lt;input name="reject" type="button" value="Reject" /&gt;
  &lt;/form&gt;
  &lt;/div&gt;
  </code></pre>
  
  <p>when i clicked detailed on my select * database, it show popup like above, when it popup it offer button to approve and reject...
  how to process that button, so it could update my database status...
  something like this : </p>
  
  <pre><code>if(isset($_POST['approved'])) {
    $sql = "UPDATE `kb_tiny` SET `status` = 'approved'";
  </code></pre>
  
  <p>please help me how supposed the code to work... 
  the data already showing but approved button and reject not working :(</p>