title: Show bootstrap popover after submitting form that is in modal
body: <p>I have a form, that is in remote modal window. Here is the modal in index.html:</p>
  
  <pre><code>&lt;a data-toggle="modal"  href="contact.html" data-target="#myModal"&gt;Contact Me&lt;/a&gt;
  &lt;div class="modal fade" id="myModal" tabindex="-1" role="dialog" aria-labelledby="myModalLabel" aria-hidden="true"&gt;
      &lt;div class="modal-dialog"&gt;
          &lt;div class="modal-content"&gt;
          &lt;/div&gt;
      &lt;/div&gt;
  &lt;/div&gt;
  </code></pre>
  
  <p>Than here is acctual form in contact.html:</p>
  
  <pre><code>&lt;script&gt;
      $('#pop').popover({content: 'Message send.'},'click');    
  &lt;/script&gt;
  
  &lt;form id="contactform" method="post" action="data/send_mail.php"&gt;
      &lt;!--
      here is whole form content
      --&gt;
      &lt;button type="submit" value="Submit" class="btn btn-custom" data-toggle="popover" data-placement="right" id="pop"&gt;Send&lt;/button&gt;
  &lt;/form&gt;
  </code></pre>
  
  <p>Now my <strong>problem is:</strong> I need to show bootstrap popover after not clicking on the submit button but when the form is really send. I do the validation in browser(no php regular expressions or jQuery validate, simply in browser).
  My approach was following (after gathering the data from from and creating message in send_mail.php):</p>
  
  <pre><code>@mail($email_to, $email_subject, $email_message, $headers);
      /* Here should be something that will make my popover show, like: echo ('$('#pop').popover('show')');*/
  sleep(2);
  echo "&lt;meta http-equiv='refresh' content=\"0; url=../index.html\"&gt;";
  </code></pre>
  
  <p>But this second command (in comment does not work). I am total newbie in php, so I try to make my solutions as simple as possible. I hope that I have explained my problem properly.
  Thx...</p>