alterando a raiz do aplicativo para o aplicativo mvc spring no tomcat
Estou trabalhando com um exemplo de recurso RESTEasy 2.0 no Spring MVC 3.0 e de acordo com o Tomcat 6. Posso acessar meu recurso através de http: //localhost:8080/examples-resteasy-2.1-SNAPSHOT/contacts, mas gostaria de acessar através http: // localhost: 8080 / Contacts ou mesmo http: // localhost: 8080 / myservice / Contacts
Há algo que eu precise alterar na maneira como meu aplicativo é mapeado para o caminho?
Web.xml
<web-app>
<servlet>
<servlet-name>springmvc</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>classpath:springmvc-servlet.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>springmvc</servlet-name>
<url-pattern>/contacts/*</url-pattern>
</servlet-mapping>
</web-app>
springmvc-servlet.xml
<beans xmlns="http: //www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 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
http: //www.springframework.org/schema/beans http: //www.springframework.org/schema/beans/spring-beans.xsd
">
<context:component-scan base-package="org.jboss.resteasy.examples.springmvc" />
<context:annotation-config />
<import resource="classpath:springmvc-resteasy.xml" /> <!-- this is included in the resteasy-spring library-->
<bean id="viewResolver"
class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<property name="viewClass" value="org.springframework.web.servlet.view.JstlView" />
<property name="prefix" value="/WEB-INF/" />
<property name="suffix" value=".jsp" />
</bean>
</beans>
minha classe de recurso RESTEasy
@Controller
@Path("/contacts")
public class ContactsResource {
...