使用Jetty以编程方式部署servlet

Question

我有一个servlet,我希望使用Jetty以编程方式部署.servlet使用Spring,它的web.xml指向Spring上下文XML文件,正如您所期望的那样.

目前,我只是尝试Jetty文档中的示例代码,但使用自己的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();

这导致以下异常:

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

如何设置我的servlet以便Jetty知道它的web.xml和Spring上下文在哪里？

真的很感激任何帮助!

编辑

所以,显然我不需要web.xml但我确实需要将Jetty指向我的Spring上下文.我尝试过以下内容:

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

但它不起作用(仍然产生相同的例外).我也尝试在ServletHolder上设置"contextConfigLocation"无济于事.

Answer 1

对于那些感兴趣的人,我按照以下方式工作:

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();

我失踪的关键步骤是

context.addEventListener(new ContextLoaderListener());

它加载Spring上下文.