Jetty: Pass object from main method to servlet

Solution 1:

Embedded Jetty is so wonderful here.

You have a few common options:

1. Direct instantiation of the servlet, use constructors or setters, then hand it off to Jetty via the ServletHolder (can be any value or object type)
2. Add it to the ServletContext in your main, and then access it via the ServletContext in your application (can be any value or object type).

Examples:

package jetty;

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.eclipse.jetty.server.Server;
import org.eclipse.jetty.servlet.ServletContextHandler;
import org.eclipse.jetty.servlet.ServletHolder;

public class ObjectPassingExample
{
    public static void main(String args[]) throws Exception
    {
        Server server = new Server(8080);

        ServletContextHandler context = new ServletContextHandler();
        context.setContextPath("/");

        // Option 1: Direct servlet instantiation and ServletHolder
        HelloServlet hello = new HelloServlet("everyone");
        ServletHolder helloHolder = new ServletHolder(hello);
        context.addServlet(helloHolder, "/hello/*");

        // Option 2: Using ServletContext attribute
        context.setAttribute("my.greeting", "you");
        context.addServlet(GreetingServlet.class, "/greetings/*");

        server.setHandler(context);
        server.start();
        server.join();
    }

    public static class HelloServlet extends HttpServlet
    {
        private final String hello;

        public HelloServlet(String greeting)
        {
            this.hello = greeting;
        }

        @Override
        protected void doGet(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException
        {
            resp.setContentType("text/plain");
            resp.getWriter().println("Hello " + this.hello);
        }
    }

    public static class GreetingServlet extends HttpServlet
    {
        private String greeting;

        @Override
        public void init() throws ServletException
        {
            this.greeting = (String) getServletContext().getAttribute("my.greeting");
        }

        @Override
        protected void doGet(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException
        {
            resp.setContentType("text/plain");
            resp.getWriter().println("Greetings to " + this.greeting);
        }
    }
}

Solution 2:

Singleton

You want to pass the same single instance to each servlet?

Use the Singleton pattern to create a single instance that is available globally.

The simplest fool-proof way to do that in Java is through an Enum. See Oracle Tutorial. Also see this article and the book Effective Java: Programming Language Guide, Second Edition (ISBN 978-0-321-35668-0, 2008) by Dr. Joshua Bloch.

So no need to pass an object. Each servlet can access the same single instance through the enum.

Per web app

If you want to do some work when your web app is first launching but before any servlet in that web app has handled any request, write a class that implements the ServletContextListener interface.

Mark your class with the @WebListener annotation to have your web container automatically instantiate and invoke.