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brisi
servlet responds wiyh status 4004
1 Answer(s)      2 years and 5 months ago
Posted in : Ajax


function getXMLObject()  //XML OBJECT
{
   var xmlHttp = false;
   try {
     xmlHttp = new ActiveXObject("Msxml2.XMLHTTP")  // For Old Microsoft Browsers
   }
   catch (e) {
     try {
       xmlHttp = new ActiveXObject("Microsoft.XMLHTTP")  // For Microsoft IE 6.0+
     }
     catch (e2) {
       xmlHttp = false   // No Browser accepts the XMLHTTP Object then false
     }
   }
   if (!xmlHttp && typeof XMLHttpRequest != 'undefined') {
     xmlHttp = new XMLHttpRequest();        //For Mozilla, Opera Browsers
   }
   return xmlHttp;  // Mandatory Statement returning the ajax object created
}

var xmlhttp = new getXMLObject(); //xmlhttp holds the ajax object

function ajaxFunction() {
  if(xmlhttp) { 
   var txtname = document.getElementById("txtname");
    xmlhttp.open("POST","http://localhost:8080/Admin/web/getname",true); //getname will be the servlet name
    xmlhttp.onreadystatechange  = handleServerResponse;
    xmlhttp.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded');
    xmlhttp.send("txtname=" + txtname.value); //Posting txtname to Servlet
  }
}

function handleServerResponse() {
   if (xmlhttp.readyState == 4) {
       alert("hyn");
       alert(xmlhttp.status);
     if(xmlhttp.status == 200) {
       document.myForm.message.innerHTML=xmlhttp.responseText; //Update the HTML Form element 
     }
     else {
        alert("Error during AJAX call. Please try again");
     }
   }
}


SERVLET
import java.io.*;

import java.text.*;

import java.util.*;

import javax.servlet.*;

import javax.servlet.http.*;

public class getname extends HttpServlet {

 public void init(ServletConfig config) throws ServletException { 
  super.init(config);
 }

 public void destroy() {

 }

 public void doPost(HttpServletRequest request,HttpServletResponse response) throws IOException, ServletException {
  String name = null;
  PrintWriter out = response.getWriter();
  if(request.getParameter("txtname") != null) {
   name = request.getParameter("txtname");
  }
                else {
                       name = "";
                }
           out.println("You have successfully made Ajax Call:" + name);
 }
}

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View Answers

February 7, 2012 at 12:41 PM


It seems that the server is unable to find your jsp or servlet file. Either you have specified wrong path to run a jsp or servlet or may be you have saved the file with different name.

If you are running a servlet, servlet mapping is essential. Check it.



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