一 需求
在登录成功的页面，显示登录成功的人名
二 方案
1 实现方法
通过Request对象实现
2 实现代码
2.1 LoginAction
package com.cakin.actions;
//这是一个action（表示小队长，需要继承Action）
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import org.apache.struts.action.Action;
import org.apache.struts.action.ActionForm;
import org.apache.struts.action.ActionForward;
import org.apache.struts.action.ActionMapping;
import com.cakin.forms.UserForm;
public class LoginAction extends Action {
//我们需要重新编写一个方法：execute会被自动调用，有点类似servlet的service方法。
@Override
public ActionForward execute(ActionMapping mapping, ActionForm form,
HttpServletRequest request, HttpServletResponse response)
throws Exception {
//把form转成对应的UserForm对象
UserForm userForm=(UserForm)form;
//简单验证
if("123".equals(userForm.getPassword())){
//把名字存放到request对象域
request.setAttribute("username", userForm.getUsername());
return mapping.findForward("ok");
}
else{
return mapping.findForward("err");
}
}
}
2.2 wel.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 'wel.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>
welcome <%=request.getAttribute("username").toString() %> <br>
<a href="/strutslogin/">返回重新登录</a>
</body>
</html>
三 测试结果