filter过滤器实现验证跳转_返回验证结果

1. 需求背景

需要对某个请求url进行拦截，模拟是否可以进入某一个接口，如果拦截需要返回数据false，别问我为何不用intercept拦截器。

2. web.xml

<filter>    <filter-name>restfulFilter</filter-name>    <filter-class>com.jeenotes.utils.filter.RestfulFilter</filter-class> 过滤器路径   
</filter>    <filter-mapping>    <filter-name>restfulFilter</filter-name>    <url-pattern>/aaa/*</url-pattern>  aaa表示拦截的url,如果你想拦截所有，直接/*即可。   
</filter-mapping> 

 

3. 自定义的Filter

 public class RestfulFilter implements Filter {   

    private AAAService aaaService;@Overridepublic void init(FilterConfig filterConfig) throws ServletException {}@Overridepublic void doFilter(ServletRequest request, ServletResponse response, FilterChain chain)throws IOException, ServletException {
　　　　 //由于filter 优先级要高，所以直接@Autowired引入service是不存在的
        //如下是HttpServletRequest req = (HttpServletRequest)request;HttpServletResponse resp = (HttpServletResponse)response;ServletContext sc = req.getSession().getServletContext();
        //如下是创建service过程XmlWebApplicationContext cxt = (XmlWebApplicationContext)WebApplicationContextUtils.getWebApplicationContext(sc);//aaaServiceImpl 是aaaService实现类if(cxt != null && cxt.getBean("aaaServiceImpl") != null && aaaService == null)aaaService = (AAAService) cxt.getBean("aaaServiceImpl");//此处是逻辑if(成功){chain.doFilter(request, response); //进入请求的url}else{req.getRequestDispatcher("/xxx某某url").forward(request,response);//跳转自己指定的url}}@Overridepublic void destroy() {}}

要跳转的/xxx某某url

@RequestMapping(value = "/getEntranceStatus", method = RequestMethod.GET, produces = "text/html;charset=UTF-8")public String getEntranceStatus(HttpServletRequest request){//此处就是返回一个false}