Struts2的Session超時返回登入頁面
今天做一個S2SM的專案時，在session超時或者失效時返回到登陸頁面重新登陸，前臺使用的easyui框架
1、首先修改web.xml
<session-config>
<session-timeout>1</session-timeout>
</session-config>
2、自定義一個登入超時的攔截器
package com.tungkong.util;
import java.util.Hashtable;
import javax.servlet.http.HttpServletResponse;
import org.apache.struts2.ServletActionContext;
import com.opensymphony.xwork2.ActionInvocation;
import com.opensymphony.xwork2.interceptor.AbstractInterceptor;
@SuppressWarnings("serial")
public class LoginedCheckInterceptor extends AbstractInterceptor {
/** session過期、登入有效性及操作的許可權驗證攔截器 */
@SuppressWarnings("unchecked")
@Override
public String intercept(ActionInvocation ai) throws
Exception {
//取得請求的URL
String url = ServletActionContext.getRequest().getRequestURL().toString();
HttpServletResponse response = ServletActionContext.getResponse();
response.setHeader("Pragma", "No-cache");
response.setDateHeader("Expires", 0);
response.setHeader("Cache-Control", "no-cache"
);response.setHeader("Cache-Control", "no-store");
response.setHeader("Content-Type", "text/html; charset=UTF-8");
response.setHeader("Cache-Control", "no-store");
response.setHeader("Content-Type", "text/html; charset=UTF-8");
Hashtable<String, Object> userinfo = null;
//對登入與登出請求直接放行，不予攔截
if (url.indexOf("loginAction") != -1 || url.indexOf("logoutAction") != -1) {
return ai.invoke();
} else {
//驗證Session是否過期
if (!ServletActionContext.getRequest().isRequestedSessionIdValid()) {
//session過期,轉向session過期提示頁,最終跳轉至登入頁面
return "tologin";
} else {
//驗證是否已經登入
userinfo = (Hashtable<String, Object>) ServletActionContext.getRequest().getSession().getAttribute("userSessionHt");
if (userinfo == null) {
//尚未登入,跳轉至登入頁面
return "tologin";
} else {
return ai.invoke();
}
}
}
}
}
3、在struts.xml中配置攔截器，並宣告一個全域性的result，當session失效的時候攔截器會轉發到登陸頁面，由於我使用的是多個struts檔案，故應該這麼設定
<?xml version="1.0" encoding="UTF-8" ?>
<!DOCTYPE struts PUBLIC
"-//Apache Software Foundation//DTD Struts Configuration 2.0//EN"
"http://struts.apache.org/dtds/struts-2.0.dtd">
<struts>
<!-- 載入預設的 struts2 配置檔案 -->
<include file="struts-default.xml" />
<!-- 業務模組配置 -->
<include file="struts-userinfo.xml" />
………………
<constant name="struts.i18n.encoding" value="UTF-8" />
<constant name="struts.objectFactory" value="spring"></constant>
<package name="tksm" extends="struts-default">
<interceptors>
<interceptorname="loginedCheck"class="com.tungkong.util.LoginedCheckInterceptor"/>
<interceptor-stack name="mystack">
<interceptor-ref name="loginedCheck" />
</interceptors><interceptor-ref name="defaultStack" />
</interceptor-stack>
<global-results>
<result name="exception">/exception.jsp</result>
<result name="tologin">/relogin.jsp</result>
</global-results>
<global-exception-mappings>
<exception-mapping exception="java.lang.Exception" result="exception" />
</global-exception-mappings>
</package>
</struts>
4、在struts-userinfo.xml中設定
<?xml version="1.0" encoding="UTF-8" ?>
<!DOCTYPE struts PUBLIC
"-//Apache Software Foundation//DTD Struts Configuration 2.0//EN"
"http://struts.apache.org/dtds/struts-2.0.dtd">
<struts>
<package name="tksm-user" extends="tksm">
<default-interceptor-ref name="mystack" />
<action name="loginAction" class="userAction" method="login">
<result name="success">/WEB-INF/index/main.jsp</result>
<result name="error">/WEB-INF/index/error.jsp</result>
</action>
……………………
</package>
</struts>
5、新建relogin.jsp頁面
由於頁面巢狀在iframe下，跳轉時需要跳轉到其父頁面，因此加個中間的jsp，攔截器配置跳轉到此頁面，再由此頁面跳轉到登入頁面。
relogin.jsp
<%@ page language="java" import="java.util.*" pageEncoding="UTF-8"%>
<%
String path = request.getContextPath();
String basePath = request.getScheme()+"://"+request.getServerName()+":"+request.getServerPort()+path+"/";
%>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
<head>
<base href="<%=basePath%>">
<title>My JSP 'index.jsp' starting page</title>
<meta http-equiv="pragma" content="no-cache">
<meta http-equiv="cache-control" content="no-cache">
<meta http-equiv="expires" content="0">
<meta http-equiv="keywords" content="keyword1,keyword2,keyword3">
<meta http-equiv="description" content="This is my page">
<!--
<link rel="stylesheet" type="text/css" href="styles.css">
-->
</head>
<body>
<script type="text/javascript">
alert("對不起！您長時間對系統未進行操作，登入已超時，請重新登入！");
window.top.location.href="<%=basePath%>login.jsp";
</script>
</body>
</html>
當我們登陸後一分鐘不做任何操作重新整理後則會跳轉到登陸頁面
這篇部落格主要來源於：
http://blog.csdn.net/java_cxrs/article/details/5519743
http://blog.sina.com.cn/s/blog_a72f208a01014gha.html
感謝兩位！