Open a form action url page in a popup box when only a specific submit button is clicked

Question

I have a form with two submit buttons (send message & update status). I want to open a popup window when send_message button clicked and display the send_message form. Please advice how to proceed

<form name="send_messages" action="/sendMessage/" method="post" accept-charset="UTF-8"> 
     <div class="user-add"> 
               <span> 
             <select name="user-status" id="user-status" onchange="showDiv(this)"> 
         <option value="">Select Status</option> 
         <option value="0">Enable</option> 
         <option value="1">Disable</option> 
     </select></span> 
         <button id="hidden_div" style="display:none;" name="button-status" value="user-status-send" 
                 onchange="this.form.submit()"><i class="fa fa-pencil statusup"></i>Update Status 
         </button> 
         <span><button id="merge_button" name="button-status" value="message-send"  > 
                <i class="fa fa-envelope"> </i>Send Message</button> 
         </span> 
   </div> 

Answer 1

您可以在模式<iframe id='submitFormFrame' />创建框架，然后将target属性设置为指向该框架<form target='#submitFormFrame'>