filter过滤器实现验证跳转_返回验证结果

Posted 2020-11-02 niceyoo

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
        
  }