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);
}