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).

the best question i ever seen. nice adverb "shortest code". – Amir Oct 18 '22 at 17:58 — Amir, Oct 18 '22 at 17:58

score 14 · Accepted Answer · edited Feb 11 '12 at 10:44

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.

Jonathan Holloway · Answer 2 · 2009-06-20T18:11:14.343

3

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.

edited Jun 20 '09 at 18:11

answered Jun 20 '09 at 17:43

Jonathan Holloway

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I don't see the spring tag on the OP question. Why throw a 800 pound gorilla like spring at something so simple?! – philk Apr 17 '20 at 07:07

Janus Troelsen · Answer 3 · 2013-07-26T23:28:43.570

Works with Jetty 8:

import org.eclipse.jetty.server.Server;
import org.eclipse.jetty.webapp.WebAppContext;

public class Main {
    public static void main(String[] args) throws Exception {
            Server server = new Server(8080);
            WebAppContext handler = new WebAppContext();
            handler.setResourceBase("/");
            handler.setContextPath("/");
            handler.addServlet(new ServletHolder(new MyApp()), "/*");
            server.setHandler(handler);
            server.start();
    }
}

score 2 · Answer 4 · answered Apr 05 '15 at 10:53

I've written a library, EasyJetty, that makes it MUCH easier to embed Jetty. It is just a thin layer above the Jetty API, really light-weight.

Your example would look like this:

import com.athaydes.easyjetty.EasyJetty;

public class Sample {

    public static void main(String[] args) {
        new EasyJetty().port(80).servlet("/*", MyApp.class).start();
    }

}

score 1 · Answer 5 · answered Sep 12 '13 at 00:43

1

        Server server = new Server(8080);
        Context root = new Context(server, "/");
        root.setResourceBase("./pom.xml");
        root.setHandler(new ResourceHandler());
        server.start();

answered Sep 12 '13 at 00:43

sendon1982

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Shortest code to start embedded Jetty server

5 Answers5

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