jsp页面中获取servlet请求中的参数的办法详解

在JAVA WEB应用中，如何获取servlet请求中的参数，并传递给跳转的JSP页面？例如访问http://localhost:8088/bbs?id=1

当执行这个bbs servlet时，将url参数id的值传递给bbs.jsp页面？

1.首先要配置web.xml，见下面的配置：

 bbs
 
  org.openjweb.core.servlet.BBSServlet
 
 
 
 bbs
 /bbs
 

2.编写servlet类：

package org.openjweb.core.servlet;

import java.io.IOException;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
 
public class BBSServlet extends HttpServlet 
{
 private static final long serialVersionUID = 1L;
  public BBSServlet() 
  {
    super();
    // TODO Auto-generated constructor stub
  }
  
 protected void doGet(HttpServletRequest request, HttpServletResponse response) 
 throws ServletException, IOException 
 { 
 　//http://bbs.csdn.net/topics/90438353
   request.setCharacterEncoding("UTF-8");  //设置编码
   String id = request.getParameter("id");
   request.setAttribute("id", id);
   request.getRequestDispatcher("/bbs.jsp").forward(request, response);
 }
 
 protected void doPost(HttpServletRequest request, HttpServletResponse response) 
 throws ServletException, IOException 
 {
 doGet(request, response);
 
 }
}

在应用根目录创建bbs.jsp文件，内容为：

 <%@ page contentType="text/html;charset=UTF-8"%>
 <%
out.println(request.getAttribute("id"));

 %>

注意很多人传递参数不成功是因为是在doGet方法中调用doPost,这里doGet方法不要调用doPost.