IE, tomcat 4.0 - running servlet kicks to localhost7070 search screen

I moved my shopping cart website (my first) to a PC running xp, downloaded j2sdk1.4.1_02 & tomcat4.0, set classpath, java_home, catalina_home. Set html in the webapps directory of Tomcat, java servlets in the webapps/mywebsite/WEB-INF/classes folder and recompiled sucessfuly. Tomcat runs on localhost7070, I can see the HTML in /mywebsite but when I click on the 'submit button' to run the servlet, my IE momentarily shows 'not found' but immediately goes to an ATT Worldnet site with a search field filled with localhost:7070.
I have no idea of what setting might be off on IE but I have checked the directorys and system variables multiple times and can't figure this out. Any ideas would be greatly appreciated.
Bruce

Your web.xml (in WEB-INF) have your Servlet URL mapping. Try and see if you can get in contact with your Servlet manualy by typing in URLs like
HTTP://localhost:7070/servlet/ProductLookup
HTTP://localhost:7070/mywebsite/servlet/ProductLookup
HTTP://localhost:7070/mywebsite/ProductLookup
and when you find it, compare it the the URL in your HTML, relative to the URL used load the HTML page.

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Tomcat crashes while running servlet chat in IE

Hi all!
I've seen similar problems posted about three years ago, but I didn't see an answer for it.
I'd be very grateful if you could help me.
I'm writing a chat, the code was taken from the J.Hunter "Servlet programming book" O'reilly, "absurdly simple chat", and adjusted (I only need the Http version). It's an applet-servlet chat.
When executed in IE, Tomcat crashes after a few (usually 5) sent messages. When executed in a debugger (I use Forte, the last available version, 4 update 1) the program works just fine...
Another question is about debugging an applet, executed in a browser - how do I do this?
Here is the code for the servlet:
=================================
import java.io.*;
import java.net.*;
import java.util.*;
import javax.servlet.*;
import javax.servlet.http.*;
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public void doGet(HttpServletRequest req, HttpServletResponse res)
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res.setContentType("text/plain");
PrintWriter out = res.getWriter();
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out.println(getNextMessage());
// doPost() accepts a new message and broadcasts it to all
// the currently listening HTTP and socket clients.
public void doPost(HttpServletRequest req, HttpServletResponse res)
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// Accept the new message as the "message" parameter
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// Broadcast it to all listening clients
if (message != null) broadcastMessage(message);
// Set the status code to indicate there will be no response
res.setStatus(res.SC_NO_CONTENT);
// getNextMessage() returns the next new message.
// It blocks until there is one.
public String getNextMessage() {
// Create a message sink to wait for a new message from the
// message source.
return new MessageSink().getNextMessage(source);
// broadcastMessage() informs all currently listening clients that there
// is a new message. Causes all calls to getNextMessage() to unblock.
public void broadcastMessage(String message) {
// Send the message to all the HTTP-connected clients by giving the
// message to the message source
source.sendMessage(message);
// MessageSource acts as the source for new messages.
// Clients interested in receiving new messages can
// observe this object.
class MessageSource extends Observable {
public void sendMessage(String message) {
setChanged();
notifyObservers(message);
// MessageSink acts as the receiver of new messages.
// It listens to the source.
class MessageSink implements Observer {
String message = null; // set by update() and read by getNextMessage()
// Called by the message source when it gets a new message
synchronized public void update(Observable o, Object arg) {
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message = (String)arg;
// Wake up our waiting thread
notify();
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// Tell source we want to be told about new messages
source.addObserver(this);
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try { wait(); } catch (Exception ignored) { }
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source.deleteObserver(this);
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// But first set the message instance variable to null
// so update() and getNextMessage() can be called again.
String messageCopy = message;
message = null;
return messageCopy;
=============================
The code for the applet is
=============================
import java.applet.*;
import java.awt.*;
import java.io.*;
import java.net.*;
import java.util.*;
public class TryChatApplet extends Applet implements Runnable {
TextArea text;
Label label;
TextField input;
Thread thread;
String user;
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// If so, explain to the user that this applet needs to be loaded
// from a server in order to communicate with that server's servlets.
URL codebase = getCodeBase();
if (!"http".equals(codebase.getProtocol())) {
System.out.println();
System.out.println("*** Whoops! ***");
System.out.println("This applet must be loaded from a web server.");
System.out.println("Please try again, this time fetching the HTML");
System.out.println("file containing this servlet as");
System.out.println("\"http://server:port/file.html\".");
System.out.println();
System.exit(1); // Works only from appletviewer
// Browsers throw an exception and muddle on
// Get this user's name from an applet parameter set by the servlet
// We could just ask the user, but this demonstrates a
// form of servlet->applet communication.
user = getParameter("user");
if (user == null) user = "anonymous";
// Set up the user interface...
// On top, a large TextArea showing what everyone's saying.
// Underneath, a labeled TextField to accept this user's input.
text = new TextArea();
text.setEditable(false);
label = new Label("Say something: ");
input = new TextField();
input.setEditable(true);
setLayout(new BorderLayout());
Panel panel = new Panel();
panel.setLayout(new BorderLayout());
add("Center", text);
add("South", panel);
panel.add("West", label);
panel.add("Center", input);
public void start() {
thread = new Thread(this);
thread.start();
String getNextMessage() {
String nextMessage = null;
while (nextMessage == null) {
try {
URL url = new URL(getCodeBase(), "/servlet/TryChatServlet");
HttpMessage msg = new HttpMessage(url);
InputStream in = msg.sendGetMessage();
DataInputStream data = new DataInputStream(
new BufferedInputStream(in));
nextMessage = data.readLine();
catch (SocketException e) {
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System.out.println("Can't connect to host: " + e.getMessage());
try { Thread.sleep(5000); } catch (InterruptedException ignored) { }
catch (FileNotFoundException e) {
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try { Thread.sleep(5000); } catch (InterruptedException ignored) { }
catch (Exception e) {
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System.out.println("General exception: " +
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try { Thread.sleep(1000); } catch (InterruptedException ignored) { }
return nextMessage + "\n";
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while (true) {
text.appendText(getNextMessage());
public void stop() {
thread.stop();
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message = user + ": " + message; // Pre-pend the speaker's name
try {
URL url = new URL(getCodeBase(), "/servlet/TryChatServlet");
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Properties props = new Properties();
props.put("message", message);
msg.sendPostMessage(props);
catch (SocketException e) {
// Can't connect to host, report it and abandon the broadcast
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catch (FileNotFoundException e) {
// Servlet doesn't exist, report it and abandon the broadcast
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catch (Exception e) {
// Some other problem, report it and abandon the broadcast
System.out.println("General exception: " +
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switch (event.id) {
case Event.ACTION_EVENT:
if (event.target == input) {
broadcastMessage(input.getText());
input.setText("");
return true;
return false;
=====================================
HttpMessage
======================================
import java.io.*;
import java.net.*;
import java.util.*;
public class HttpMessage {
URL servlet = null;
String args = null;
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out.close();
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StringBuffer buf = new StringBuffer();
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buf.append(URLEncoder.encode(name) + "=" + URLEncoder.encode(value));
if (names.hasMoreElements()) buf.append("&");
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Those files are the only files needed to execute the program.

The whole Tomcat crashes, and no exception
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<servlet-class>
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thanks in advance,
ann

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do I config the web.xml file?

Take a look at:
http://java.sun.com/dtd/web-app_2_3.dtd
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type Exception report
message Internal Server Error
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exception
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     at org.apache.catalina.core.ContainerBase.invoke(ContainerBase.java:943)
     at org.apache.catalina.core.StandardContext.invoke(StandardContext.java:2347)
     at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:180)
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     at org.apache.catalina.valves.ErrorDispatcherValve.invoke(ErrorDispatcherValve.java:170)
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     at org.apache.catalina.core.StandardPipeline.invokeNext(StandardPipeline.java:566)
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     at org.apache.catalina.loader.WebappClassLoader.loadClass(WebappClassLoader.java:1254)
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     at org.apache.catalina.core.ContainerBase.invoke(ContainerBase.java:943)
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     at org.apache.catalina.core.ContainerBase.invoke(ContainerBase.java:943)
     at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:174)
     at org.apache.catalina.core.StandardPipeline.invokeNext(StandardPipeline.java:566)
     at org.apache.catalina.core.StandardPipeline.invoke(StandardPipeline.java:472)
     at org.apache.catalina.core.ContainerBase.invoke(ContainerBase.java:943)
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Hi,
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Thanks

To start, make sure you have specified the servlet and mapping in your application's WEB-INF/web.xml file.

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Hi My Friends
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Hi ,
could someone please help me.
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please help

hi
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<url-pattern>/the ref name which u want call servlet</url-pattern>
<servlet-mapping>
Here the servlet name sholud be same .
The url pattern name must match with the action attribute value in html.
now you place your html file in root folder.
place your servlet class file in classes folder.
start ur browser type this http://localhost:8080/htmlfilename.

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I lost my one day to solve this problem but at the end of the day i got mad
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Tomcat: 6
i have directory structure as
C:\Program Files\Apache Software Foundation\Tomcat 6.0\webapps
now i have create my own directory to run my own servlets its has structure like
C:\Program Files\Apache Software Foundation\Tomcat 6.0\webapps\myApp\WEB-INF
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WEB-INF has two sub directories named classes and src and one file web.xml
in src i have Simple.java servlet
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my classpath variable has value
C:\Program Files\Apache Software Foundation\Tomcat 6.0\lib\servlet-api.jar
my path variable has value
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<?xml version="1.0" encoding="ISO-8859-1"?>
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PUBLIC "-//Sun Microsystems, Inc.//DTD Web Application 2.3//EN"
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<description>My Test Servlet </description>
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<servlet-name>Simple</servlet-name>
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HTTP Status 404 - /myApp/Simple
type Status report
message /myApp/Simple
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Apache Tomcat/6.0.16
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if i recall, you need to put that class in a package
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can't run ,it give me a error than http:404 not found.
could you tell me how i can solve the problem.
Thank u very much.

Hi,
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Regards

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Find Lomboz J2EE plugin and a tutorial of how to use it :)

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<body>
</html>
servlet file
servletcalled.java contains
public void doPost(HttpServletRequest req,HttpServletResponse res ){
java.io.PrintWriter out = req.getWriter();
out.println("Hi, executed");
i put the callservlet.html in webapps/examples/ and servletcalled.class was in webapps/examples/Web-inf/classes/
After starting the tomcat and running the program html file is getting exceuted but when i click on the submit button this error is prompted
type Status report
message servletcalled.class
description The requested resource (servletcalled.class) is not available.
Thanks in advance

Thanks,
I created a new directory in webapps
s "webapps/test".
Test directory contains
1. callservlet.html file
2. another directory Web-inf (i.e,
webapps/test/Web-inf)
Web-inf directory contains
1. web.xml file
2. another directory classes (i.e,
webapps/test/Web-inf/calsses)
classes directory contains
1. servletcalled.class file
web.xml file contains
<?xml version="1.0" encoding="ISO-8859-1"?>
<!DOCTYPE web-app
PUBLIC "-//Sun Microsystems, Inc.//DTD Web
Web Application 2.3//EN"
"http://java.sun.com/dtd/web-app_2_3.dtd">
<web-app>
<servlet>
<servlet-name>Example</servlet-name>
<servlet-class>Example</servlet-class>
</servlet>
</web-app>
-->should be: <servlet-class>servletcalled</servlet-called>
then in the <web-app> scope define this:
<servlet-mapping>
<servlet-name>Example</servlet-name>
<url-pattern>/servlets/Example</url-pattern>
</servlet-mapping>
>
thanks in advance i am working hard on that but
notable to get the solutionbtw by reading the documentation of tomcat and tutorials of jave on the java site (here), you would have known this!

How to run servlets under Tomcat 6 in java1.5

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i am unable to run servlets under tomcat 6,i checked environment variables ....all are correct..im getting http 503 error....

A 503 response says
503 Service Unavailable
The server is currently unable to handle the request due to a
temporary overloading or maintenance of the server
You apparently have the server mis-configured. Recheck the configuration (or reinstall Tomcat per its instructions.)

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