spring mvc 有如下两个映射，应该怎么处理这两个映射呢？

Question

@RequestMapping(value="/users/{id}") public String renderUser(@PathVariable("id") int id){    //do something}@RequestMapping(value="/users/{id}.{format}") public String renderUserOf(@PathVariable("id") int id,@PathVariable("format") String format){    //do something}第一个 "/users/{id}" 能够正常处理但是第二个 /users/{id}.{format} 却不能正常解析比如 /users/7 会被映射为 "/users/{id}" ,但 /users/7.json 却也被映射为了 "/users/{id}"

Answer 1

Spring根据后缀来决定编组格式.
如果你想返回json.

<bean class="org.springframework.web.servlet.view.ContentNegotiatingViewResolver">
                <!-- 设置为false开启对Accept Header的支持-->
                <property name="ignoreAcceptHeader" value="false"/>
                <!-- 在没有扩展名时的默认形式 -->
                 <property name="defaultContentType" value="application/xml"/>        
       
        <property name="mediaTypes">
            <map>
                <entry key="json" value="application/json" />
                <entry key="xml" value="application/xml" />
                <entry key="jsp" value="text/html"/>
                <entry key="do" value="text/html"/>
            </map>
        </property>
        <property name="viewResolvers">
            <list>          
                <bean class="org.springframework.web.servlet.view.InternalResourceViewResolver">
                    <property name="viewClass" value="org.springframework.web.servlet.view.InternalResourceView"/>
                    <property name="prefix" value="/WEB-INF/jsp/"></property>              
                </bean>
                <bean class="org.springframework.web.servlet.view.BeanNameViewResolver" />
             
            </list>
        </property>
        <property name="defaultViews">
            <list>            
                <!-- for application/xml -->           
                <bean class="org.springframework.web.servlet.view.xml.MarshallingView" >
                    <property name="marshaller">
                        <ref bean="castorMarshaller"/>
                    </property>
                </bean> 
                <!-- for application/json -->  
                <bean class="org.springframework.web.servlet.view.json.MappingJacksonJsonView" />                
            </list>
        </property>
    </bean>

请求id.xml就得到xml格式的,请求id.json就是json格式的~没有后缀名会根据Accept类型.