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

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：

<servlet>
 <servlet-name>bbs</servlet-name>
 <servlet-class>
  org.openjweb.core.servlet.bbsservlet
 </servlet-class>
 </servlet>
 <servlet-mapping>
 <servlet-name>bbs</servlet-name>
 <url-pattern>/bbs</url-pattern>
 </servlet-mapping>

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.