I'm sure this is a noob question, and I've spent the better part of an hour trawling stackoverflow for an answer but nobody seems to have my case so here we go...
I have a new webapp that uses Spring MVC. Most of the app (99%) is pure REST, so it doesn't have a "view" as such but rather simply sends JSON back down the wire, or sends an alternate HTTP Status for errors etc.
The exception is the login page which needs to be an actual JSP, but somehow the configuration I am using to map my REST controllers is leaving me in a state where normal JSP mappings fail.
Here's what I've got:
In my dispatcher servlet config, the relevant portions are:
<bean class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<property name="prefix" value="/WEB-INF/pages/"/>
<property name="suffix" value=".jsp"/>
</bean>
In my attempts to get it working, I have also added a mapping to the "HomeController" which currently just redirects to my login JSP:
<bean name="/" class="com.somepackage.HomeController"/>
Now, in the web.xml I have:
<servlet>
<servlet-name>spring-dispatcher</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>spring-dispatcher</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
<listener>
<listener-class>
org.springframework.web.context.ContextLoaderListener
</listener-class>
</listener>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>
/WEB-INF/spring-dispatcher-servlet.xml
</param-value>
</context-param>
This works fine for my RESTful controllers, which look like this:
#Controller
#RequestMapping(value = "/api/user")
public class BlahBlahController {...
My "HomeController", which just looks like this:
#Controller
#RequestMapping(value = "/")
public class HomeController extends AbstractController {
#Override
protected ModelAndView handleRequestInternal(HttpServletRequest request, HttpServletResponse response) throws Exception {
return new ModelAndView("login");
}
}
IS triggered when I hit the "/" url, but I get this error in the logs:
WARNING: No mapping found for HTTP request with URI [/WEB-INF/pages/login.jsp] in DispatcherServlet with name 'spring-dispatcher'
Now I get what it's saying, it doesn't know how to resolve /WEB-INF/pages/login.jsp (this page does exist btw), but I'm stuck as to how I need to alter things to get this to work.
I'm a little confused on how it's supposed to work. Anyone got any clues?
Thanks.
OK.. I found the answer, it's the url-pattern in the dispatcher config.
Instead of:
<servlet-mapping>
<servlet-name>spring-dispatcher</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
It should be
<servlet-mapping>
<servlet-name>spring-dispatcher</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
I had actually found this answer elsewhere and tried it but "thought" it wasn't working, then realized the reason I thought this was unrelated to the root cause.
No idea why this would work and the other wouldn't.. but one problem at a time...
Put RequestMapping at the method level, I tried at my code and it worked. You don't need to define HomeController bean if you are using #Controller and having a proper "context:component-scan"
#Override
#RequestMapping(value = "/")
protected ModelAndView handleRequestInternal(HttpServletRequest request, HttpServletResponse response) throws Exception {
return new ModelAndView("login");
}
you can also use below code if you just want to redirect a login view for all "/" access.
<mvc:view-controller path="/" view-name="login"/>
Checkout the mvc show case project from github for a helpful reference.
Related
Im new to spring MVC and REST.. I'm having an issue with a simple test controller I've put together from example I've found here and from the spring docs..
When I hit the url http://localhost:8080/test-api/user/14 I get the error below
Im getting the error:
Sep 23, 2015 11:26:55 AM org.springframework.web.servlet.PageNotFound noHandlerFound
WARNING: No mapping found for HTTP request with URI [/test-api/user/14] in DispatcherServlet with name 'testapi'
Im using xml to config.. Im not ready to move to java config.
web.xml
Spring Web MVC Application
<servlet>
<servlet-name>springtest</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>springtest</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/testapi-servlet.xml</param-value>
</context-param>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
testapi-servlet.xml - only contains the component scan and annotation driven elements
<context:component-scan base-package="com.springtest.testapi" />
<mvc:annotation-driven />
SpringTest.java
package com.springtest.testapi.api;
#RestController
public class SpringTest {
#RequestMapping(value="/user/{id}", method = RequestMethod.GET)
public User getUser(#PathVariable int id) {
User u = new User(id,"Test","Me");
return u;
}
What handler should I be defining.. None of the examples or docs state that a handler needs to be defined..
Remove the contextConfigLocation.
Replace the following:
<servlet-mapping>
<servlet-name>springtest</servlet-name>
<url-pattern>/test-api</url-pattern>
</servlet-mapping>
Make sure your xml file is test-api-servlet.xml and not testapi-servlet.xml
I found my issue. I mispelled the package in the component scan. The sample code I have was edited so It didn't fully represent what I had and was actually correct.
I'm trying to add atmosphere with REST inside a JBoss apllication. The intention is to keep the web application running and add a Web Service. (Deployed application via Ant)
I have added the following jars:
atmosphere-runtime.jar
jaxrs-api-1.1-RC2.jar
jboss-seam-resteasy-2.2.0.GA.jar
resteasy-jaxrs-1.1-RC2.jar
web.xml:
<servlet>
<description>AtmosphereServlet</description>
<servlet-name>AtmosphereServlet</servlet-name>
<servlet-class>org.atmosphere.cpr.AtmosphereServlet</servlet-class>
<load-on-startup>0</load-on-startup>
<async-supported>true</async-supported>
</servlet>
<servlet-mapping>
<servlet-name>AtmosphereServlet</servlet-name>
<url-pattern>/atmosphere/*</url-pattern>
</servlet-mapping>
atmosphere.xml:
<atmosphere-handlers>
<!-- RESTEasy -->
<atmosphere-handler support-session="false"
context-root="/*/atmosphere"
class-name="org.atmosphere.handler.ReflectorServletProcessor"
interceptorClasses="org.atmosphere.interceptor.IdleResourceInterceptor">
<property name="servletClassName"
value="org.jboss.seam.servlet.SeamResourceServlet"/>
</atmosphere-handler>
</atmosphere-handlers>
My java class like this:
#Path("/test")
public class Foo {
#Context HttpServletRequest req;
#GET
#Path("/suspend")
public Response suspend() {
/*
AtmosphereResource r = (AtmosphereResource)
req.getAttribute("org.atmosphere.cpr.AtmosphereResource");
r.resumeOnBroadcast(true);
r.setBroadcaster(BroadcasterFactory.getDefault().lookup("/test", true))
.suspend();
*/
return Response.ok().build();
}
}
URL: http://XXX.XXX.XX.XXX:8080/posmngr/atmosphere/test
Everytime I try to enter the URL I get the following error:
ERROR [[AtmosphereServlet]] Servlet.service() for servlet AtmosphereServlet threw exception
org.atmosphere.cpr.AtmosphereMappingException: No AtmosphereHandler maps request for /posmngr/atmosphere/test/
I changed the root-context to "/atmosphere/*" but I get a 404.
Am I missing something? Any help would be really appreciated. Thanks in advance!
I was a litle trickier due to our own framework. I had to integrate atmosphere with spring and seam and consider the order that those are initialized.
My atmosphere.xml context-root now is: "/atmosphere/*"
I removed the atmosphere servlet definition from web.xml and placed in an external file that is consumed by spring.
Web.xml fragment:
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>classpath*:META-INF/servlet/*.xml</param-value>
</context-param>
and the external file contains:
<beans default-lazy-init="true">
<bean id="atmosphereHandler" class="com.ourcompany.webapp.support.ServletUrlHandlerMapping" parent="abstractHandler">
<property name="mappings">
<props>
<prop key="/atmosphere/**">atmosphereResourceController</prop>
</props>
</property>
</bean>
<bean id="atmosphereResourceController" class="org.springframework.web.servlet.mvc.ServletWrappingController" depends-on="servletContextInitializer">
<property name="servletClass" value="org.atmosphere.cpr.AtmosphereServlet"/>
</bean>
</beans>
No changes were made to the java and the efective URL is: http://XXX.XXX.XX.XXX:8080/posmngr/atmosphere/rest/test/suspend
Thanks a lot!
I am trying to develp a rest service using apache camel. My project is a spring mvc war deployed on tomcat.
I dont want to use apache cxf (cxf servlet).
public class SampleRouter extends RouteBuilder {
#override
public void configure() throws Exception {
from("cxfrs://http://localhost:1234/sample")
.process (new Processor() {
public void process(Exchange exchange) throws Exception {
System.out.println("test");
}
})).setBody(constant("SUCCESS"));
}
}
#Path("/sample")
public class SampleResource {
#GET
public void test() {
}
}
web.xml has dispatcherservlet, contextloaderlistener.
dispatcher-servlet.xml has mvc:annotation-drivem, context:component-scan,
<camelContext id="server" trace="true" xmlns="http://camel.apache.org/schema/spring">
<contextScan />
</camelContext>
pom.xml has camel-core, camel-cxf, camel-stream, cxf-rt-transports-http-jetty, cxf-rs-frontend-jaxrs, camel-spring, spring-webmvc, spring-web, spring-context.
Tomcat runs on 8080, there seems to be no exception when server comes up. But, I tried hitting the url (http://localhost:1234/sample), nothing seems to be happening.
What am i missing? I would eventually extend this to REST to Spring DSL or REST to Java DSL with authentication, filters and interceptors.
I also tried cxf:rsServer and referred that in router class.
Also, in the future if i have to use https instead of http? or how do i have the url not hard-coded?
It may be too late, but to consume HTTP requests, one may use Apache Camel Servlet component
http://camel.apache.org/servlet.html
You need to setup the resourceClass option on the cxfrs endpoint. Here is an example
from("cxfrs://http://localhost:1234/sample?resourceClasses=my.pachage.SampleResource")
You can find some example in camel-cxfrs component page.
If you want to export a CXF service through servlet transport, you need to do some work as it said.
If you want to change the address dynamically, you can take look at the camel properties component.
If you are looking to start a camel route by a consuming cxf rest service which uses the servlet transport then you need to :
Clean up your pom.xml and remove any references to jetty.
Add the CXF servlet to your web.xml
<servlet>
<servlet-name>CXFServlet</servlet-name>
<servlet-class>org.apache.cxf.transport.servlet.CXFServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<!-- all our webservices are mapped under this URI pattern -->
<servlet-mapping>
<servlet-name>CXFServlet</servlet-name>
<url-pattern>/services/*</url-pattern>
</servlet-mapping>
Add the servlet-transport dependency:
<dependency>
<groupId>org.apache.cxf</groupId>
<artifactId>cxf-rt-transports-http</artifactId>
<version>${cxf-version}</version>
</dependency>
In your spring/camel configuration
<import resource="classpath:META-INF/cxf/cxf.xml"/>
<import resource="classpath:META-INF/cxf/cxf-servlet.xml" />
<cxf:rsServer id="rsServer" address="/restService"
serviceClass="com.something.test.SimpleServiceImpl"
loggingFeatureEnabled="true" loggingSizeLimit="20" />
Build a route from this consumer endpoint as:
from("cxfrs:bean:rsServer?bindingStyle=SimpleConsumer")
.to("log:TEST?showAll=true")
You can now view/(invoke with a method) the endpoint using : http://host:port/context/services/restService?_wadl
I am trying to get resteasy working on JBoss AS6 Final (SEAM 2 app), but I cant seem the get the most basic example working, as I understand it, resteasy should be ready to go, I have tried the following example from here but the urls simply result in 404 errors with no response
package uk.co.rest.test;
import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.core.Application;
public class Library extends Application {
#GET
#Path("/books")
public String getBooks() {
System.out.println("Check");
return "done";
}
}
with the following added to my web.xml
<context-param>
<param-name>resteasy.servlet.mapping.prefix</param-name>
<param-value>/rest</param-value>
</context-param>
<servlet>
<servlet-name>resteasy-servlet</servlet-name>
<servlet-class>org.jboss.resteasy.plugins.server.servlet.HttpServletDispatcher</servlet-class>
<init-param>
<param-name>javax.ws.rs.Application</param-name>
<param-value>uk.co.rest.test.Library</param-value>
</init-param>
</servlet>
<servlet-mapping>
<servlet-name>resteasy-servlet</servlet-name>
<url-pattern>/rest/*</url-pattern>
</servlet-mapping>
I get the feeling that the resteasy.deployer that is bundled with JBoss is not doing its job, but im not sure how to go about debugging it
Any help would be great im pulling my hair out over this one!!
RESTEasy must be configured in order to be exposed as a service. You can do it either directly or via Seam's resource servlet.
To use RESTEasy directly, I find the easiest way is configuring it as a Servlet filter. There's little to do other than adding the filter to your web.xml as documented in http://docs.jboss.org/resteasy/docs/2.3.0.GA/userguide/html/Installation_Configuration.html#filter.
When using Seam this is however unnecessary, as Seam is capable of deploying RESTEasy services via its resource servlet quite simply (documented in http://docs.jboss.org/seam/2.2.0.GA/reference/en-US/html/webservices.html#d0e22093). You first declare RESTEasy's application component like this:
<resteasy:application resource-path-prefix="/rest" />
And create your providers that will be automatically deployed into the configured path, for example:
#Name("libraryService")
#Path("/library")
public class Library implements Serializable {
#In(create=true) private transient BookHome bookHome;
#GET #Path("/{book}")
#Produces("text/plain")
public String getBooks(#PathParam("book") String id) {
bookHome.setId(id);
return bookHome.getInstance().getTitle();
}
}
You can then access the RESTEasy service via:
http://localhost:8080/yourapp/seam/resource/rest/library/1
The advantage of going the Seam way is mostly ease of use. You do need to include an extra jar: jboss-seam-resteasy.jar.
You seem to have misunderstood the role of javax.ws.rs.Application. Your Library class does not have to extend javax.ws.rs.Application in order to expose the getBooks()-method.
Create a class that extends javax.ws.rs.Application. Override the getSingletons()-methods and return a set of instances that has methods you wish to expose:
public class MyApplication extends javax.ws.rs.Application {
#Override public Set<Object> getSingletons(){
return Collections.<Object>singleton(new Library());
}
}
In your web.xml, change the javax.ws.rs.Application init-param, so that it points to the MyApplication class.
Im using annotation driven Spring WS 2.0.2 to create a simple Webservice, but the enpoint mapping was not found.
Input and Output are jdom Elements to keep it as simple as possible.
The Webservice is running with Java 1.6 on Tomcat 6.0.29 wich returns an error
page (The requested Resource () is not available) to my SoapUI Service Test.
Here is the Error I get in my logging:
WARNING: No endpoint found for [SaajSoapMessage (http://foo.bar/myTest)myRequest]
Here are the parts of the configuration I deem relvant for the Endpoint mapping:
(If there are more relevant parts I am missing please ask back...)
Schema (WEB-INF/xsd/myTest.xsd)
targetNamespace="http://foo.bar/myTest"
...
<element name="myRequest" type="tns:string"/>
<element name="myResponse" type="tns:string"/>
web.xml (WEB-INF/web.xml)
<servlet-class>
org.springframework.ws.transport.http.MessageDispatcherServlet
</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/spring/config.xml</param-value>
</init-param>
<init-param>
<param-name>transformWsdlLocations</param-name>
<param-value>true</param-value>
</init-param>
Spring config (/WEB-INF/spring/config.xml)
<sws:annotation-driven/>
<sws:dynamic-wsdl id="myTest"
portTypeName="myTest"
localUri="/"
targetNamespace="http://foo.bar/myTest">
<sws:xsd location="/WEB-INF/xsd/myTest.xsd"/>
</sws:dynamic-wsdl>
Endpoint (src/main/java/bar/foo/MyEndpoint.java)
#Endpoint
public class MyEndpoint{
#PayloadRoot(localPart="myRequest",namespace="http://foo.bar/myTest")
#ResponsePayload
public Element mySearch( #RequestPayload Element myRequest){
return myRequest;
}
}
Searching for a sollution I found it contained in this answer
Adding
...
xmlns:context="http://www.springframework.org/schema/context"
...
xsi:schemaLocation=" ...
http://www.springframework.org/schema/context
http://www.springframework.org/schema/context/spring-context-3.0.xsd ... "
<context:component-scan base-package="bar.foo"/>
to my Spring configuration let the servlet find my Endpoint.
My problem was, that no sample code in a spring documentation I found contained
this step and its relevance.
Well - actually I found this code snipplet in a tutorial earlier, but it was a bit overloaded with features I did not need, and as in the official docs it was not explained why it was necessary.