判断用户session 是否过期，如果过期，让用户重新登录

野性酷女 2023-06-25 02:19 17阅读 0赞

1、在web.xml配置：

WmsFilter com.test.wms.filter.WmsFilter true WmsFilter /mgr/\*/\*基于请求路径过滤\*/

2、对应过滤器代码：

public class WmsFilter implements Filter \{  
// 日志  
Logger logger = LoggerFactory.getLogger(WmsFilter.class);

public void destroy() \{  
// TODO Auto-generated method stub  
\}

public void doFilter(ServletRequest sreq, ServletResponse sresp, FilterChain chain)  
throws IOException, ServletException \{  
if (!(sreq instanceof HttpServletRequest) || !(sresp instanceof HttpServletResponse)) \{  
throw new ServletException(“OncePerRequestFilter just supports HTTP requests”);  
\}  
HttpServletRequest httpRequest = (HttpServletRequest) sreq;  
HttpServletResponse httpResponse = (HttpServletResponse) sresp;  
httpResponse.setHeader(“Cache-Control”,“no-cache”);  
httpResponse.setHeader(“Pragma”,“no-cache”);  
httpResponse.setDateHeader (“Expires”, -1);  
httpResponse.setHeader(“P3P”,“CP=CAO PSA OUR”);

HttpSession session = httpRequest.getSession();  
if(!httpResponse.isCommitted())\{  
if(session != null)\{  
Object object = session.getAttribute(“userName”);  
String userName = object == null ? null : (String) object;  
if (userName == null) \{  
boolean isAjaxRequest = isAjaxRequest(httpRequest);  
if (isAjaxRequest) \{  
httpResponse.sendError(HttpStatus.UNAUTHORIZED.value(), “您已经太长时间没有操作,请刷新页面”);  
\}  
httpResponse.sendRedirect("/wms/index.jsp");  
\}  
\}else\{  
httpResponse.sendRedirect("/wms/index.jsp");  
\}  
\}  
chain.doFilter(sreq, sresp);  
\}

/\*\*

*  判断是否为Ajax请求
 *  
 *  @param request
 *      HttpServletRequest
 *  @return 是true, 否false  
    \*/  
    public static boolean isAjaxRequest(HttpServletRequest request) \{  
    return (request.getHeader(“X-Requested-With”) != null  
    && “XMLHttpRequest”.equals(request.getHeader(“X-Requested-With”).toString()));  
    \}

public void init(FilterConfig config) throws ServletException \{  
\}  
\}