jsp页面中获取servlet请求中的参数的办法详解
发布时间:2020-05-23 21:29:40 所属栏目:Java 来源:互联网
导读:在JAVAWEB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/bbsid=1
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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: <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,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. 您可能感兴趣的文章:
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