JBoss Version: 4.2.3GA. This works in apache tomcat 6.0. In JBoss, I had to add the following setting: -Dorg.apache.catalina.STRICT_SERVLET_COMPLIANCE=false
to get the forward to work, but now when I load the page I get the error below. It feels like I am doing this in a way JBoss doesn't like, but I haven't seen any other examples. Does anyone know a way to get this to work?
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 "";
}
The exception:
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)
After getting some help from another location, I realized that I was connecting my JSP and restlet in a funny way, and what I really wanted to do was use a Viewable. This also works way better in JBoss. Here is a summary of what I ended up with:
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);
}
If you are not using Jersey, you may want to do this:
@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));
}
}
来源:https://stackoverflow.com/questions/6984338/how-to-forward-from-a-jax-rs-service-to-jsp