现在有两个jsp页面文件,一个是index.jsp,一个是userList,都存放在WebContent目录下(Eclipse的根目录),要求编写一个servlet名为GetUserListServlet.java,完成从index.jsp中的点击某一按钮跳转到userList.jsp,并将点击人的姓名也显示在userList.jsp页面中, 1. 采用servlet2.0规范编写如下: public class GetUserListServlet extends HttpServlet { protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { this.doPost(request, response); } protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { String hit_name = ""; //假设hit_name作为参数传回userList.jsp页面 request.setAttribute("hitName", ___1___); //跳转至指定页面 String url = "userList.jsp"; //RequestDispatcher完成请求转发 RequestDispatcher dispatcher = request.getRequestDispatcher(____2___); dispatcher.forward(request,response); } } 2. web.xml中注册此GetUserListServlet,注册内容如下: This is the description of my J2EE component This is the display name of my J2EE component GetUserListServlet cn.edu.bitc.servlet.GetUserListServlet GetUserListServlet /getHitName 3. 现在index.jsp中有一表单提交按钮,代码片断如下:

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4.此时如果能顺利跳转到userList.jsp,相要在页面中显示出点击人的姓名,userList.jsp的关键代码片断如下: <%@ page language="java" contentType="text/html;charset=utf-8" pageEncoding="utf-8"%> <%@ taglib prefix="c" uri="http://java.sun.com/jsp/jstl/core" %>