Shortest code to start embedded Jetty server

Question

I'm writing some example code where an embedded Jetty server is started. The server must load exactly one servlet, send all requests to the servlet and listen on localhost:80

My code so far:

static void startJetty() {
        try {
            Server server = new Server();

            Connector con = new SelectChannelConnector();
            con.setPort(80);
            server.addConnector(con);

            Context context = new Context(server, "/", Context.SESSIONS);
            ServletHolder holder = new ServletHolder(new MyApp());
            context.addServlet(holder, "/*");

            server.start();
        } catch (Exception ex) {
            System.err.println(ex);
        }

    }

Can i do the same with less code/lines ? (Jetty 6.1.0 used).

Answer 1

static void startJetty() {
        try {
            Server server = new Server();
            Connector con = new SelectChannelConnector();
            con.setPort(80);
            server.addConnector(con);
            Context context = new Context(server, "/", Context.SESSIONS);
            context.addServlet(new ServletHolder(new MyApp()), "/*");
            server.start();
        } catch (Exception ex) {
            System.err.println(ex);
        }
    }

Removed unnecessary whitespace and moved ServletHolder creation inline. That's removed 5 lines.

Answer 2

You could configure Jetty declaratively in a Spring applicationcontext.xml, e.g:

http://roopindersingh.com/2008/12/10/spring-and-jetty-integration/

then simply retrieve the server bean from the applicationcontext.xml and call start... I believe that makes it one line of code then... :)

((Server)appContext.getBean("jettyServer")).start();

It's useful for integration tests involving Jetty.