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Editorial Team
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Asked: 2026-05-23T18:34:13+00:00

I am trying to get an XML response from a servlet. The servlet returns

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I am trying to get an XML response from a servlet. The servlet returns a content type of “application/xml”. Using XmlHttpRequest, I can get responseText, but not responseXml. I was wondering if this had something to do with the content type or the request type (I’m doing a GET)…?

Many thanks!

I’ve pared down all my files. I think I’m setting that all correctly. Here’s what I have:

——- HTML ——

<html>
<head>
<title></title>
<script src="js/jboard_simple.js" type="text/javascript"> </script>
</head>
<body>

    <div id="myDiv">
        <h2>No results yet....</h2>
    </div>

    <form name="searchForm" id="searchForm_id">
        <input type="text" name="searchString" id="searchString_id" />
        <button type="button" onclick="loadXMLDoc()">Perform Search</button>
    </form>

</body>
</html>

——- JavaScript ———–

function loadXMLDoc() {
    document.getElementById("myDiv").innerHTML = "searching...";

    var xmlhttp;
    if (window.XMLHttpRequest) {
        xmlhttp = new XMLHttpRequest();
    } else {
        xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
    }
    xmlhttp.onreadystatechange = function() {

        if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
            // Do something here...
            alert(xmlhttp.responseText);
            alert(xmlhttp.responseXml);

            processSearchServletResponse(xmlhttp.responseText);
        }
    }

    // Find teh search string
    var searchString_el = window.document.getElementById('searchString_id');
    var searchString = searchString_el.value;
    alert('searchString: ' + searchString);

    var searchUrl = "/SimpleServlet?searchString=" + searchString;

    xmlhttp.open("GET", searchUrl, true);
    xmlhttp.send();
}

function processSearchServletResponse(xmlTxt) {
    document.getElementById("myDiv").innerHTML = xmlTxt;    
}

——- Servlet ——–

import java.io.BufferedOutputStream;
import java.io.IOException;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

import org.apache.log4j.Logger;
import org.apache.log4j.LogManager;

public class SimpleServlet extends HttpServlet {
    private static final long serialVersionUID = 1L;

    private static Logger logger = LogManager.getRootLogger();

    public void service(HttpServletRequest req, HttpServletResponse res) throws ServletException, IOException {
        BufferedOutputStream bs = null;
        String simpleResponse = "<?xml version=\"1.0\" encoding=\"UTF-8\"?><root>hi</root>";


        try {
            res.setContentType("text/xml");
            res.setCharacterEncoding("UTF-8");

            bs = new BufferedOutputStream(res.getOutputStream());
            bs.write(simpleResponse.getBytes());

        } catch (Exception ex) {
            logger.error("JboardSearchServlet.service(): error = ", ex);
        } finally {

            bs.flush();
            bs.close();
        }
    }
}
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1 Answer

  1. Editorial Team

    Oh dear! I just figured out my whole problem (after someone else looked at it).

    I was using ‘responseXml’ rather than ‘responseXML’. Darn those upper case letters =)

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