如何在struts2中登录后跳转到前置页面

package actions; 

import javax.servlet.http.HttpServletRequest; 

import javax.servlet.http.HttpSession; 

import org.apache.struts2.ServletActionContext; 

import com.opensymphony.xwork2.ActionInvocation; 

import com.opensymphony.xwork2.interceptor.AbstractInterceptor; 

public class AuthorizationInterceptor extends AbstractInterceptor 

{ 

private static final long serialVersionUID = 2071704648277594728L; 

public String intercept(ActionInvocation invocation) throws Exception    

     {     

         HttpServletRequest request=ServletActionContext.getRequest();    

         HttpSession session=request.getSession();    

//通过session 判断用户是否登录

if(session!=null && session.getAttribute("user")!=null) {    

return invocation.invoke();    

          }    

//设置客户原来请求的url地址  

         session.setAttribute("url", getGoingURL(request)); 

return Action.LOGIN;    

     }    

private String getGoingURL(HttpServletRequest request)    

     {     

         String url=request.getHeader("Referer");    

if(url==null || url.equals(""))    

         {    

//如果请求的是ACTION 则通过该代码获取URL地址,当然可以完善其功能

             url = request.getContextPath()+request.getServletPath()+"?"+request.getQueryString(); 

         }    

return url; 

     } 

} 

参考于：http://blog.csdn.net/oyjqdlp364356378/article/details/5715227