springmvc 使用RESTFul风格的URL

Love The Way You Lie 2022-05-27 09:51 214阅读 0赞

web.xml配置

<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*:spring/spring-mvc.xml</param-value>  
            </init-param>  
            <load-on-startup>1</load-on-startup>  
        </servlet>  
      
        <servlet-mapping>  
            <servlet-name>springmvc</servlet-name>  
            <url-pattern>*.do</url-pattern>  
        </servlet-mapping>  
        
        
    
        <servlet>  
            <servlet-name>springRESTful</servlet-name>  
            <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>  
            <init-param>  
                <param-name>contextConfigLocation</param-name>  
                <param-value>classpath*:spring/spring-mvc.xml</param-value>  
            </init-param>  
        </servlet>  
      
        <servlet-mapping>  
            <servlet-name>springRESTful</servlet-name>  
            <url-pattern>/</url-pattern>  
        </servlet-mapping>

springmvc.xml配置

<mvc:resources location="/js/" mapping="/js/**"/>
    	<mvc:resources location="/css/" mapping="/css/**"/>
    	<mvc:resources location="/images/" mapping="/images/**"/>

接着在ACTION里配置

//id跟是把页面传来的数据匹配绑定到参数上 
    	@RequestMapping("/gUserInfo/{id}")
    	public @ResponseBody TUserinfo gUserinfo(@PathVariable("id") String user) throws Exception {
    		TUserinfo u = userService.findByUserName(user);
    		
    		return u;
    	}

页面上直接访问xxx/gUserInfo/id就可以了