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  • 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; @Override public void init(FilterConfig filterConfig) throws ServletException { } @Override public 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 } } @Override public 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
            
      }
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  • 原文地址:https://www.cnblogs.com/niceyoo/p/8979748.html
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