登陆后跳转到之前页面

一、非集群

  在非集群的情况下，我们一般用session来管理用户的，一般的框架都会有拦截器或安全处理，在此我只写个大致方法和思路，具体的类就不写了，

 

 

	
		public byte[] doLogin(HttpServletRequest req) {
			StringBuilder builder = new StringBuilder();
			Enumeration result = req.getAttributeNames();
			
			//获取该请求的参数为还原做准备
			while (result.hasMoreElements()) {
				String attri = (String) result.nextElement();
					builder.append(attri).append("=").append(
							req.getAttribute(attri).toString()).append(",");//此处用","隔开每个参数，用&时在页面会出错

			}

			String returnUrl = req.getRequestURL() + "?" + builder.toString();
			
			req.setAttribute("returnUrl", returnUrl);
		}

 

 

 

 

login页面

 

<form id="logForm" method="post" action="login">
 //此处记录将要登陆后的将要返回的页面url
 <input type="hidden" name="returnUrl" value="${ param.returnUrl }"/>



 <div>用户名：<input type="text" maxlength="25" name="userName" id="userName" /></div>
      	<div>密　码：<input type="password" maxlength="50" name="password" id="password" /></div>
      	<div id="buttonBar">
          <input value="确定" type="submit" />
          <div class="clear"></div>
        </div>
      </form>
      <div id="info"> </div>
    </div>
    <script language="javascript">
	<!--
	(function(){
		$("#logForm").form({
			userName:{
				cls:"largerInput",
				validate:function(){
					return $(this).val().length>2;
				},
				validateText:"请输入3-8个字符"
			},
			password:{
				cls:"largerInput",
				validate:function(){
					return $(this).val().length>4;
				},
				validateText:"请输入5-10个字符"
			}
		});
		
	})();
	-->
	</script> 

 

登陆后跳转

 

//之前的登录判断没写，执行到这步时已经登录上了
		public void Login(HttpServletRequest req,HttpServletResponse resp){
			
				String returnUrl = (String) req.getAttribute("returnUrl");
				
				if(null != returnUrl || !returnUrl.equals("")){
					returnUrl = returnUrl.replace(",", "&");
					resp.sendRedirect(returnUrl);
				}else{
					resp.sendRedirect(req.getContextPath() + "/home");
				}
		}