I have form as follows, it require to sent an action to my java Servlet to do an update to the database.
How do I submit the form without the page get reloaded here? 
Currently with action="myServlet" it keep direct me to a new page. And if I remove the action to myServlet, the input is not added to my database.
<form name="detailsForm" method="post" action="myServlet" 
      onsubmit="return submitFormAjax()">
    name: <input type="text" name="name" id="name"/> <br/>
    <input type="submit" name="add" value="Add" />
</form>
In the view of my Java servlet, request.getParameter will look for the name and proceed to add it into my db.
@Override
protected void doPost(HttpServletRequest request, HttpServletResponse response) 
  throws ServletException, IOException
{   
    if (request.getParameter("add") != null) {
        try {
            Table.insert(name);
        } catch (Exception ex) {
            ex.printStackTrace();
        }
    }
}
In my JavaScript part, I have a submitFormAjax function
function submitFormAjax()
{
    var xmlhttp;
    if (window.XMLHttpRequest) {
        // code for modern browsers
        xmlhttp = new XMLHttpRequest();
    } else {
        // code for IE6, IE5
        xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
    }
    xmlhttp.onreadystatechange = function() {
        if (xmlhttp.readyState == 4 && xmlhttp.status == 200)
            alert(xmlhttp.responseText); // Here is the response
        }
    var id = document.getElementById("name").innerHTML;
    xmlhttp.open("POST","/myServlet",true);
    xmlhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
    xmlhttp.send("name=" + name); 
}
 
     
     
     
    