java - Spring RESTful web service using only XML config -


i've been using xml configuration make mvc web applications.( no annotation)

now want make restful web service spring not find tutorial doesn't use annotation.

is there way build restful web service xml configuration ?
or have use annotation ?

for example, can deploy mvc pattern web application using xml configuration below.

 <bean class="org.springframework.web.servlet.view.internalresourceviewresolver">       <property name="prefix" value="/web-inf/jsp/" />       <property name="suffix" value=".jsp" />    </bean>     <bean class="org.springframework.web.servlet.mvc.multiaction.parametermethodnameresolver" id="springparametermethodnameresolver">     <property name="paramname" value="action"/>    </bean>     <bean class="org.springframework.web.servlet.handler.simpleurlhandlermapping">     <property name="mappings">        <map>             <entry key="/test.do" >                 <ref bean="testcontroller"  />             </entry>             <entry key="/rest/test">                 <ref bean="testrestcontroller"/>             </entry>         </map>      </property>    </bean>     <!-- beans -->    <bean id="testmethodnameresolver" class="com.rhcloud.riennestmauvais.spring.test.testmethodnameresolver">    </bean>     <!-- test -->    <bean class="com.rhcloud.riennestmauvais.spring.test.testcontroller" id="testcontroller">         <property name="delegate" ref="testdelegate"/>         <property name="methodnameresolver" ref="testmethodnameresolver"></property>         <!-- <property name="methodnameresolver" ref="springparametermethodnameresolver"></property> -->    </bean>    <bean class="com.rhcloud.riennestmauvais.spring.test.testdelegate" id="testdelegate">    </bean> 

however, hit wall when trying map method url example http method : post, url : /student/1/adam - add student.
url format this. /[resource]/[id]/[name]

i map /student/1/adam controller putting pattern in entry key how should parse uri within controller ?

i parse uri using string.split() or i'm wondering if there isn't solution avoid reinventing wheel.

<?xml version="1.0" encoding="utf-8"?>    	<beans xmlns="http://www.springframework.org/schema/beans"  	 xmlns:context="http://www.springframework.org/schema/context"  	 xmlns:mvc="http://www.springframework.org/schema/mvc" xmlns:xsi="http://www.w3.org/2001/xmlschema-instance"  	 xmlns:p="http://www.springframework.org/schema/p"  	 xsi:schemalocation="  		http://www.springframework.org/schema/beans              http://www.springframework.org/schema/beans/spring-beans-4.0.xsd          http://www.springframework.org/schema/context          http://www.springframework.org/schema/context/spring-context-4.0.xsd          http://www.springframework.org/schema/mvc          http://www.springframework.org/schema/mvc/spring-mvc-4.0.xsd">             		<context:component-scan base-package="com.apmc.rest" />   		<mvc:annotation-driven />   		      </beans>

this rest-servlet.xml. file must configure in web.xml using dispatcherservlet class

<servlet>   	<servlet-name>rest</servlet-name>   	<servlet-class>    		org.springframework.web.servlet.dispatcherservlet   	</servlet-class>   	<load-on-startup>2</load-on-startup>   </servlet>     <servlet-mapping>   	<servlet-name>rest</servlet-name>   	<url-pattern>/rest/*</url-pattern>   </servlet-mapping>

above code write in web.xml load-on-startup 1 give spring-security.xml , spring-config.xml