在php [关闭]按钮点击事件的弹出窗口中打开一个表单

Question

I want to open A Form in popup window on button's click event in JAVASCRIPT. can anybody help me about this??

Answer 1

php is a server side scripting language. You can do it with jquery. here a demo jsfiddle shows modal popup. Please go through it.

<!-- Button trigger modal -->
<button class="btn btn-primary btn-lg" data-toggle="modal" data-target="#myModal">
  Launch demo modal
</button>

<!-- Modal -->
<div class="modal fade" id="myModal" tabindex="-1" role="dialog" aria-labelledby="myModalLabel" aria-hidden="true">
  <div class="modal-dialog">
    <div class="modal-content">
      <div class="modal-header">
        <button type="button" class="close" data-dismiss="modal"><span aria-hidden="true">&times;</span><span class="sr-only">Close</span></button>
        <h4 class="modal-title" id="myModalLabel">Modal title</h4>
      </div>
      <div class="modal-body">
        ...
      </div>
      <div class="modal-footer">
        <button type="button" class="btn btn-default" data-dismiss="modal">Close</button>
        <button type="button" class="btn btn-primary">Save changes</button>
      </div>
    </div>
  </div>
</div>