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Home/ Questions/Q 8060783
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Editorial Team
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Editorial Team
Asked: June 5, 20262026-06-05T10:06:20+00:00 2026-06-05T10:06:20+00:00

I’ve got a servlet that I wish to deploy programmatically using Jetty. The servlet

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I’ve got a servlet that I wish to deploy programmatically using Jetty. The servlet uses Spring and it’s web.xml points to the Spring context XML file as you’d expect.

At the moment, I’m just trying the example code from the Jetty docs but with my own servlet:

Server server = new Server(8080);

ServletContextHandler context = new ServletContextHandler(ServletContextHandler.SESSIONS);
context.setContextPath("/");
server.setHandler(context);

context.addServlet(new ServletHolder(new BatchReceiver()),"/br/*");

server.start();
server.join();

This results in the following exception:

2012-05-24 14:43:20.190:INFO:oejs.Server:jetty-8.1.3.v20120416
2012-05-24 14:43:20.266:WARN:/:unavailable
java.lang.IllegalStateException: No WebApplicationContext found: no ContextLoaderListener registered?
at org.springframework.web.context.support.WebApplicationContextUtils.getRequiredWebApplicationContext(WebApplicationContextUtils.java:84)
at com.spiffymap.sealog.server.servlet.BatchReceiver.init(BatchReceiver.java:126)
at org.eclipse.jetty.servlet.ServletHolder.initServlet(ServletHolder.java:492)
at org.eclipse.jetty.servlet.ServletHolder.doStart(ServletHolder.java:312)
at org.eclipse.jetty.util.component.AbstractLifeCycle.start(AbstractLifeCycle.java:59)
at org.eclipse.jetty.servlet.ServletHandler.initialize(ServletHandler.java:778)
at org.eclipse.jetty.servlet.ServletContextHandler.startContext(ServletContextHandler.java:258)
at org.eclipse.jetty.server.handler.ContextHandler.doStart(ContextHandler.java:699)
at org.eclipse.jetty.util.component.AbstractLifeCycle.start(AbstractLifeCycle.java:59)
at org.eclipse.jetty.server.handler.HandlerWrapper.doStart(HandlerWrapper.java:90)
at org.eclipse.jetty.server.Server.doStart(Server.java:262)
at org.eclipse.jetty.util.component.AbstractLifeCycle.start(AbstractLifeCycle.java:59)
at com.spiffymap.sealog.server.servlet.TestBatchReceiver.main(TestBatchReceiver.java:26)
2012-05-24 14:43:20.335:INFO:oejs.AbstractConnector:Started SelectChannelConnector@0.0.0.0:8080

How can I set up my servlet so that Jetty knows where it’s web.xml and Spring context are?

Any help would really be appreciated!

EDIT

So, apparently I don’t need a web.xml but I do need to point Jetty to my Spring context. I’ve tried something like the following:

context.setInitParameter("contextConfigLocation", "classpath*:**/*Context.xml");

But it doesn’t work (still produces the same exception). I’ve also tried setting the “contextConfigLocation” on the ServletHolder to no avail.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-05T10:06:22+00:00Added an answer on June 5, 2026 at 10:06 am

    For those who are interested, I got this to work as follows:

    Server server = new Server(8090);
    
    ServletContextHandler context = new ServletContextHandler(ServletContextHandler.SESSIONS);
    context.setContextPath("/batch");
    
    // Setup Spring context
    context.addEventListener(new ContextLoaderListener());
    context.setInitParameter("contextConfigLocation", "classpath*:**/testContext.xml");
    
    server.setHandler(context);
    
    // Add servlets
    context.addServlet(new ServletHolder(new BatchReceiver()), "/receiver/*");
    context.addServlet(new ServletHolder(new BatchSender()), "/sender/*");       
    
    server.start();
    server.join();
    

    The key step I was missing was

    context.addEventListener(new ContextLoaderListener());
    

    which loads the Spring context.

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