ajax怎么返回上一页~~~~求解答！！！！

jsp页面的代码
String action = request.getParameter("action");
String headpage = request.getParameter("headpage");//这里是页面传回来的数据index,RamFS,index,dev......
if ("set".equals(action)) {
if (headpage != null) {
CMDHref.headpage = headpage;
out.print("{\"message\":\"ok\"}");
}
} else if ("get".equals(action)) {
headpage = CMDHref.headpage;
out.print("{\"headpage\":\""+headpage+"\"}");
}

CMDHref是一个新建立的类，类里面就2个参数
public static String headpage="";
public static String uppage="";

ajax应该不能自己返回到上一个页面吧

ajax请求完毕后客户端用js进行跳转

location.href='上一页的地址'
或者
location.href=document.referrer