Shortest code to start embedded Jetty server

问题

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).

回答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.

回答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.

回答3:

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();
    }
}

回答4:

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();
    }

}

回答5:

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

来源：https://stackoverflow.com/questions/1021971/shortest-code-to-start-embedded-jetty-server

标签

java

jetty

embedded-jetty