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How to pick up the servlet in the URL  RSS feed

Dominic Lee
Posts: 4
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Hi all, newbie again here.......

Just followed a simple excercise of how to implement a servlet. Up until this point there has been no mention of using a web.xml file!

So far I have compiled a java bit of code into a .class and placed it in my directory webapps/ch01/WEB-INF/classes as advised. The code looks like this:

import javax.servlet.*;
import javax.servlet.http.*;
public class ch01_06 extends HttpServlet
public void doGet(HttpServletRequest request,
HttpServletResponse response)
throws IOException, ServletException
PrintWriter out = response.getWriter();
out.println("A Web Page");
out.println("Hello there!");

Now the only instructions it then gives me is to use the url: http://localhost:8080/ch01/servlet/ch01_06

I then get the dreaded message this resource is not available 404 etc etc

As I have no web.xml file within my webapps folder, I am confused as to what I have to do for my servlet to be dispalyed in the browser. The instructions in the book have been quite clear up until now, can someone help?


Bear Bibeault
Author and ninkuma
Posts: 66141
IntelliJ IDE Java jQuery Mac Mac OS X
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Your tutorial is either very old, very poor, or both.

  • Servlets must be placed in a package other than the default.
  • The package hierarchy needs to reside at WEB-INF/classes, not WEB-INF.
  • You must declare and map the servelt in the deployment descriptor (web.xml).

  • I'd highly recommend finding a better and more modern tutorial.

    Michael Cropper
    Ranch Hand
    Posts: 143
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    I can highly recommend the following books which have helped my understanding massively.

    Head First Java
    Head First Java Servlets and JSP
    It is sorta covered in the JavaRanch Style Guide.
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