Getting data from servlet to ajax?

Question

Jsp / servlets seems to be much more tedious than I'd have expected it. I'm trying to call a servlet function through ajax and actually have it feed some data back to my front end , which is a jsp file.

This code returns my response as null.

This is part of my servlet. I'm trying(desperately as is fairly obvious from the code) to have it send something - anything back to ajax.

protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    String action = (String) request.getParameter("action");
if (action.equalsIgnoreCase("selectedhotel"))
    {

        response.setContentType("text/plain");  
        response.setCharacterEncoding("UTF-8"); 
        System.out.println("test");
        String attribute = (String) request.getParameter("hotel_id");
        System.out.println(attribute);
        List<Room> aRooms;
        aRooms = model.getRoomByHotel(Integer.valueOf(attribute));
        request.setAttribute("aRooms", aRooms);
        request.setAttribute("list", list);
        PrintWriter outPrintWriter = response.getWriter();
        outPrintWriter.write("ASDSADA");
            outPrintWriter.println("test");
    }

And the ajax from my JSP:

$(function(){
$("#hotelSelector li").click(function(){        
    var hid = $(this).attr('id');

    $.ajax({ type: "GET",   
         url: "AppController?action=selectedhotel&hotel_id=1",   
         success : function(text)
         {
            alert(text);
             // This will show the values. Change "alert" for $('div#mydiv').html(value) or so
         }
    });
});
});</script>

Riiight...so , please fix ?

Think of an ajax request as being no different than an ordinary http request as if you navigated to the service in your browser, because thats all it really is. — user400654
– user400654, Commented Dec 12, 2013 at 16:56
My point is build your servlet as if it were a simple hello world web page then go from there. If you're just trying to make it output something, remove all but the code that outputs something. Once you figure out how that works, start adding back piece by piece until you find what was failing. Are there any error logs that you could look at for server errors? I'm not too familiar with java so i can't really help much with possible code syntax issues. — user400654
– user400654, Commented Dec 12, 2013 at 17:06
Yea I just saw that I wasn't doing that. Changed it , still nothing. — Cata Visan
– Cata Visan, Commented Dec 12, 2013 at 17:10

Russell Shingleton · Accepted Answer · 2013-12-12 18:24:14Z

As mentioned, you really need to start small and work your way up. Get a simple "hello world" ajax response working and then tackle the more complex responses. For the more complex data response, I would recommend looking into json (see gson) to serialize the java objects in order to send back in the response writer.

First thing you should consider is using the jQuery post and get wrappers to make your life easier.

For example, your html would be like the following:

<h1>Hello: <span style="color:red" id="showMyName"></span></h1>
<form method="post" action="AjaxServlet" id="myForm">
    <input type="text" name="myName" />            
</form>
<button id="ajaxSubmit" type="submit">SEND</button>
<script type="text/javascript">
     $(document).ready(function() {
        $('#ajaxSubmit').on('click', function() {
        // To simplify things, wrap what you can in a form and serialize                      
        // it to send to the server.
            var myForm = $('#myForm');
            $.get(myForm.attr('action'), myForm.serialize(), function(data) {
                $('#showMyName').text(data);
            });
        });
    });
</script>

On the servlet side, you should start with a plain-jane servlet and once you are sure it is working, begin adding additional scope. The base servlet should be somthing like this:

// For this example, get and post will use the same base procedures.
@Override
protected void doGet(HttpServletRequest request, HttpServletResponse response)
    throws ServletException, IOException {
    processRequest(request, response);
}

@Override
protected void doPost(HttpServletRequest request, HttpServletResponse response)
    throws ServletException, IOException {
    processRequest(request, response);
}

protected void processRequest(HttpServletRequest request, 
    HttpServletResponse response) throws ServletException, IOException {
    response.setContentType("text/html;charset=UTF-8");
    PrintWriter out = response.getWriter();
    try {
        /* TODO output your response here.*/
        out.println(request.getParameter("myName"));
    } finally {
        out.close();
    }
}

One way to send a complex response is to shove all your data in a collection of some sort and use gson or some other JsonObjectMapper to convert it to a string. You can then put this String in the response writer and send it back to be parsed out by jQuery.

EDIT:

I forgot to mention that you need to make sure your servlet is being recognized by the your servlet container as well. If you haven't added the descriptor to the web.xml, it should have an entry like the following:

<servlet>
    <servlet-name>AjaxServlet</servlet-name>
    <servlet-class>org.test.AjaxServlet</servlet-class>
</servlet>
<servlet-mapping>
    <servlet-name>AjaxServlet</servlet-name>
    <url-pattern>/AjaxServlet</url-pattern>
</servlet-mapping>

vikasramireddy · Accepted Answer · 2013-12-12 17:56:41Z

0

xmlhttp.responseText is used in javascript to get the response from a servlet. how you are getting response in jquery. i think problem is in retrieving the response

edited Dec 12, 2013 at 17:56

answered Dec 12, 2013 at 17:18

vikasramireddy

881 silver badge10 bronze badges

2 Comments

Cata Visan Over a year ago

I'm not sending a null parameter since in the url of the ajax code I'm giving it an actual parameter. the if is basically compairing if ( action = "selectedhotel" ) , and it is. I also tried to add an else to it , still nothing.

vikasramireddy Over a year ago

i understood what your said.instead of action use any other variable

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