8

JBoss 版本：4.2.3GA。这适用于 apache tomcat 6.0。在 JBoss 中，我必须添加以下设置：-Dorg.apache.catalina.STRICT_SERVLET_COMPLIANCE=false让转发工作，但现在当我加载页面时，我收到以下错误。感觉就像我正在以 JBoss 不喜欢的方式这样做，但我还没有看到任何其他示例。有谁知道让这个工作的方法？

import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;

...


@GET
@Path("/forward")
public String forward(
    @Context final HttpServletRequest request,
    @Context final HttpServletResponse response) throws Exception
{
  RequestDispatcher dispatcher = request.getRequestDispatcher("/index.html");
  dispatcher.forward(request, response);
  return "";
}

例外：

java.lang.ClassCastException: $Proxy114 cannot be cast to javax.servlet.ServletRequestWrapper
    com.itt.scout.server.servlet.admin.config.ConfigController.forward(ConfigController.java:46)
    sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
    sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:39)
    sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:25)
    java.lang.reflect.Method.invoke(Method.java:597)
    com.sun.jersey.server.impl.model.method.dispatch.AbstractResourceMethodDispatchProvider$VoidOutInvoker._dispatch(AbstractResourceMethodDispatchProvider.java:151)
    com.sun.jersey.server.impl.model.method.dispatch.ResourceJavaMethodDispatcher.dispatch(ResourceJavaMethodDispatcher.java:70)
    com.sun.jersey.server.impl.uri.rules.HttpMethodRule.accept(HttpMethodRule.java:279)
    com.sun.jersey.server.impl.uri.rules.RightHandPathRule.accept(RightHandPathRule.java:136)
    com.sun.jersey.server.impl.uri.rules.ResourceClassRule.accept(ResourceClassRule.java:86)
    com.sun.jersey.server.impl.uri.rules.RightHandPathRule.accept(RightHandPathRule.java:136)
    com.sun.jersey.server.impl.uri.rules.RootResourceClassesRule.accept(RootResourceClassesRule.java:74)
    com.sun.jersey.server.impl.application.WebApplicationImpl._handleRequest(WebApplicationImpl.java:1357)
    com.sun.jersey.server.impl.application.WebApplicationImpl._handleRequest(WebApplicationImpl.java:1289)
    com.sun.jersey.server.impl.application.WebApplicationImpl.handleRequest(WebApplicationImpl.java:1239)
    com.sun.jersey.server.impl.application.WebApplicationImpl.handleRequest(WebApplicationImpl.java:1229)
    com.sun.jersey.spi.container.servlet.WebComponent.service(WebComponent.java:420)
    com.sun.jersey.spi.container.servlet.ServletContainer.service(ServletContainer.java:497)
    com.sun.jersey.spi.container.servlet.ServletContainer.service(ServletContainer.java:684)
    javax.servlet.http.HttpServlet.service(HttpServlet.java:803)
    org.jboss.web.tomcat.filters.ReplyHeaderFilter.doFilter(ReplyHeaderFilter.java:96)
4

2 回答 2

16

在从另一个位置获得一些帮助后，我意识到我正在以一种有趣的方式连接我的 JSP 和 restlet，而我真正想做的是使用 Viewable。这在 JBoss 中也更有效。以下是我最终得到的摘要：

import javax.ws.rs.core.Context;
import javax.ws.rs.Path;
import javax.ws.rs.GET;
import com.sun.jersey.api.view.Viewable;


@GET
@Path("/index")
public Viewable index(
    @Context HttpServletRequest request,
    @Context HttpServletResponse response) throws Exception
{
  request.setAttribute("key", "value");
  return new Viewable("/jsps/someJsp.jsp", null);
}
于 2011-08-09T11:52:59.747 回答
4

如果您不使用泽西岛，您可能需要这样做：

@Path("")
@ApplicationPath("blog")
@Stateless
public class BlogApplication extends Application {

    @EJB private PostEJB postEJB;

    @GET
    public void getHome(@Context HttpServletRequest request, 
                        @Context HttpServletResponse response) {
        request.setAttribute("posts", postEJB.getPosts());
        request.getRequestDispatcher("/WEB-INF/pages/blog/home.jsp")
               .forward(request, response);
    }

    @Override
    public Set<Class<?>> getClasses() {
        return new HashSet<Class<?>>(Arrays.asList(BlogApplication.class));
    }

}
于 2014-10-23T20:09:23.307 回答