带有ServletServlet的嵌入式Jetty中的错误“多个servlet映射到路径：/ *：”

Question

似乎这里有很多问题，但没有一个对我有帮助。...试图让单个Java类作为起点运行带有Jersey的嵌入式Jetty来提供网页和JSON接口...但是，即使第一步也无法提供多个页面。

这很好

ServletHolder jerseyServlet = context.addServlet(org.glassfish.jersey.servlet.ServletContainer.class, "/*");
jerseyServlet.setInitOrder(0);      
jerseyServlet.setInitParameter("jersey.config.server.provider.classnames", EntryPoint.class.getCanonicalName());

但是添加其他内容失败。 如何提供多个提供不同内容类型的页面？ 是在单个EntryPoint类中添加内容的唯一解决方案吗？

在此先感谢您提供任何提示以更改此内容

public class App {
public static void main(String[] args) throws Exception {
    ServletContextHandler context = new ServletContextHandler(ServletContextHandler.SESSIONS);
    context.setContextPath("/");
    Server jettyServer = new Server(8080);
    jettyServer.setHandler(context);

    ServletHolder jerseyServlet = context.addServlet(org.glassfish.jersey.servlet.ServletContainer.class, "/*");
    jerseyServlet.setInitOrder(0);                                                                                  
    jerseyServlet.setInitParameter("jersey.config.server.provider.classnames", EntryPoint.class.getCanonicalName());

    ServletHolder helloWorldServlet = context.addServlet(org.glassfish.jersey.servlet.ServletContainer.class, "/*");
    helloWorldServlet.setInitOrder(1);                                                                                  
    helloWorldServlet.setInitParameter("jersey.config.server.provider.classnames", HelloWorldService.class.getCanonicalName());

    try {
        jettyServer.start();
        jettyServer.join();
    } catch (Exception e){
        System.out.println("Failed running jettyServer with " + e.getMessage());
    } finally {
        jettyServer.destroy();
    }
}

}

Answer 1

居然找到了解决方案。 缺少关键信息是，您简单地需要正确的处理程序的每一个，并将它们放在处理程序列表中，瞧，到那里。

从码头文件中获取，主要是在找到它之后

public class JettyServer
{
public static void main(String[] args) throws Exception
{
    // Create a basic Jetty server object that will listen on port 8080.  Note that if you set this to port 0
    // then a randomly available port will be assigned that you can either look in the logs for the port,
    // or programmatically obtain it for use in test cases.
    Server server = new Server(8080);

    // Create the ResourceHandler. It is the object that will actually handle the request for a given file. It is
    // a Jetty Handler object so it is suitable for chaining with other handlers as you will see in other examples.
    ResourceHandler resource_handler = new ResourceHandler();
    // Configure the ResourceHandler. Setting the resource base indicates where the files should be served out of.
    // In this example it is the current directory but it can be configured to anything that the jvm has access to.
    resource_handler.setDirectoriesListed(true);
    resource_handler.setWelcomeFiles(new String[]{ "./html/index.html" });
    resource_handler.setResourceBase(".");

    //Jersey ServletContextHandler
    ServletContextHandler servletContextHandler = new ServletContextHandler(ServletContextHandler.SESSIONS);
    ServletHolder jerseyServlet = servletContextHandler.addServlet(org.glassfish.jersey.servlet.ServletContainer.class, "/api/*");
    jerseyServlet.setInitOrder(0);                                                                                  
    jerseyServlet.setInitParameter("jersey.config.server.provider.classnames", EntryPoint.class.getCanonicalName());

    // Add the ResourceHandler to the server.
    HandlerList handlers = new HandlerList();
    handlers.setHandlers(new Handler[] { resource_handler, servletContextHandler, new DefaultHandler() });
    server.setHandler(handlers);

    // Start things up! By using the server.join() the server thread will join with the current thread.
    // See "http://docs.oracle.com/javase/1.5.0/docs/api/java/lang/Thread.html#join()" for more details.
    server.start();
    server.join();
}

}

确实帮了我...

Answer 2

我认为这是您想要的：

jerseyServlet.setInitParameter("jersey.config.server.provider.classnames",
             String.join(",", Arrays.asList(EntryPoint.class.getCanonicalName(),
                     HelloWorldService.class.getCanonicalName())));

Answer 3

您没有将ServletHolder实例添加到ServletContextHandler 。

同样，两个servlet都具有相同的/*路径，我不确定，但这可能行不通，请尝试分配不同的路径并查看其是否有效。

做：

context.addServlet（jerseyServlet，“ / jersey”）;

context.addServlet（helloWorldServlet，“ / hello”）;