1.方法一 :1.1使用filter 和ajaxsetup 對ajax進行攔截並跳轉登陸頁面ajax
1 public void doFilter(ServletRequest request, ServletResponse response, FilterChain chain) 2 throws IOException, ServletException { 3 4 HttpServletResponse hresponse = (HttpServletResponse)response; 5 HttpServletRequest hrequest = (HttpServletRequest)request; 6 HttpServletResponseWrapper wrapper = new HttpServletResponseWrapper((HttpServletResponse) response); 7 8 String logonStrings = config.getInitParameter("logonStrings"); // 登陸登錄頁面 9 String redirectPath = hrequest.getContextPath() + config.getInitParameter("redirectPath");// 沒有登錄轉向頁面 10 String disabletestfilter = config.getInitParameter("disabletestfilter");// 過濾器是否有效 11 String reString = hrequest.getRequestURI(); 12 if (disabletestfilter.toUpperCase().equals("Y")) { // 過濾無效 13 chain.doFilter(request, response); 14 return; 15 } 16 // User user = ( User ) hrequest.getSession().getAttribute("user");//判斷用戶是否登陸 17 String session_key = (String) hrequest.getSession().getAttribute("token"); 18 String username = (String) hrequest.getSession().getAttribute("username"); 19 if ("".equals(session_key) || session_key == null) { 20 String[] logonList = logonStrings.split(","); 21 if (this.isContains(reString, logonList)) {// 對登陸頁面不進行過濾 22 chain.doFilter(request, response); 23 return; 24 }else{ 25 request.setAttribute("username", null); 26 boolean isAjaxRequest = this.isAjaxRequest(hrequest); 27 if (isAjaxRequest) { 28 //系統的根url 29 chain.doFilter(request, response); 30 return; 31 } 32 wrapper.sendRedirect(redirectPath); 33 return; 34 } 35 }else { 36 if(username !=null || username .equals("")) { 37 Pattern pattern = Pattern.compile("/iad/"); 38 Matcher matcher = pattern.matcher(reString); 39 Pattern pattern1 = Pattern.compile("/iad/views/login/login.jsp"); 40 Matcher matcher1 = pattern1.matcher(reString); 41 if(matcher.matches() ||matcher1.matches()){ 42 wrapper.sendRedirect("/iad/views/home/index.jsp"); 43 return; 44 }else { 45 chain.doFilter(request, response); 46 return; 47 } 48 }else{ 49 boolean isAjaxRequest = this.isAjaxRequest(hrequest); 50 if (isAjaxRequest) { 51 chain.doFilter(request, response); 52 return; 53 } 54 Pattern pattern = Pattern.compile(".*\\/views\\/login\\/login\\.jsp"); 55 Matcher matcher = pattern.matcher(reString); 56 Pattern pattern2 = Pattern.compile(".*\\/oms/"); 57 Matcher matcher2 = pattern2.matcher(reString); 58 // 是不是登錄頁面 59 if (matcher.matches() || matcher2.matches()) { 60 request.setAttribute("username", username); 61 wrapper.sendRedirect("/iad/views/home/index.jsp"); 62 return; 63 } 64 chain.doFilter(request, response); 65 return; 66 } 67 68 } 69 70 }
判斷是否爲ajaxjson
1 public static boolean isAjaxRequest(HttpServletRequest request) { 2 String header = request.getHeader("X-Requested-With"); 3 if (header != null && "XMLHttpRequest".equals(header)) 4 return true; 5 else 6 return false; 7 }
1.2 全局js代碼 能夠抽取公共js 引入到你想要的頁面session
1 $.ajaxSetup( { 2 type: "POST" , // 默認使用POST方式 3 headers: { // 默認添加請求頭 4 "Author": "CodePlayer" , 5 "Powered-By": "CodePlayer" 6 } , 7 error: function(xhr, textStatus, errorMsg){ // 出錯時默認的處理函數 8 var sessionStatus = xhr.getResponseHeader('sessionstatus'); 9 if(sessionStatus == 'timeout') { 10 var top = getTopWinow(); 11 var ids=layer.alert("導入成功!"); 12 window.location.href = "/iad/veiws/login/login.jsp"; 13 } 14 15 } 16 } );
總結: 這種方法的有個問題,就是ajax裏面的error方法是必定走的, 提醒信息會出現兩次app
方法二:使用js定時器對前臺的session進行查詢(前提是你要將這個js放到公共頁面)jsp
1 @ResponseBody 2 @RequestMapping("/loginCheck") 3 public Boolean loginCheck(HttpServletRequest request, HttpServletResponse response) { 4 5 String session_key = (String) request.getSession().getAttribute("token"); 6 String username = (String) request.getSession().getAttribute("username"); 7 if (session_key == null || username == null) { 8 response.setHeader("sessionstatus", "timeout"); 9 return false; 10 } 11 return true; 12 }
var timeid=window.setInterval("checkSession()", 1000); checkSession= function () { $.ajax({ url: "../../loginCheck", type: "POST", dataType: "json", success: function (result) { if (result != true) { window.clearInterval(timeid); Ewin.alert({message: "因爲您長時間沒有操做, session已過時, 請從新登陸"}).on(function (e) { window.location.href = "/iad/veiws/login/login.jsp"; }) } } }); };
總結:這個方法能夠將定義時間延長能夠兩秒查一次,肯定是首先1.你的系統要有公共的頁面,2.定時查詢可能致使系統變慢函數