Ajax与后台的交互

Ajax Java 交互

 

jsp代码

<%@ 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 'index.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>
<script type="text/javascript">
  var xmlHttprequest = null;
		
	function ajax(){
		if(window.ActiveXObject){
			xmlHttprequest = new ActiveXObject("Microsoft.XMLHTTP");
		
		}else if(window.XMLHttpRequest){
			xmlHttprequest = new XMLHttpRequest;
		
		}
		
		if(xmlHttprequest != null){
		
			var v1 = document.getElementById("v1").value;  
            var v2 = document.getElementById("v2").value; 
             
            
            
			xmlHttprequest.open("post","ajaxServlet.do",true);
			
			xmlHttprequest.onreadystatechange=ajaxCallBack;
			
			xmlHttprequest.setRequestHeader("Content-type",  
                "application/x-www-form-urlencoded"); 
                 
			xmlHttprequest.send("v1="+v1+"&v2="+v2);
			
		}
	
	} 
	
	function ajaxCallBack(){
		
		if(xmlHttprequest.readyState == 4){
			if(xmlHttprequest.status == 200){

				var responseText = xmlHttprequest.responseText;		
				document.getElementById("div").innerHTML = responseText;		 
			}
		
		}else{
			document.getElementById("div").innerHTML="服务器出现错误！";
		}
	}
  
  
  
  </script>
  <body>
   <input type="button" value="click here" onclick="ajax();" />  
    <input type="text" id="v1"></input>
    <input type="text" id="v2"></input>
    
    
    <div id="div"></div>
    
  </body>
</html>

　　java后台

import java.io.IOException;
import java.io.PrintWriter;

import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;


public class AjaxServlet extends HttpServlet{

	/**
	 * 
	 */
	private static final long serialVersionUID = 1733653367006626421L;

	@Override
	protected void doGet(HttpServletRequest req, HttpServletResponse resp)
			throws ServletException, IOException {
		doPost(req, resp);
	}

	@Override
	protected void doPost(HttpServletRequest req, HttpServletResponse resp)
			throws ServletException, IOException {
		PrintWriter out = resp.getWriter();
		String v1 = req.getParameter("v1");
		String v2 = req.getParameter("v2");
		
		out.print(v1+v2+"dadasdsadsa");
		out.flush();
	}

	
}

　　web.xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app version="2.5" 
	xmlns="http://java.sun.com/xml/ns/javaee" 
	xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
	xsi:schemaLocation="http://java.sun.com/xml/ns/javaee 
	http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">

  
  <servlet>
  	<servlet-name>ajax</servlet-name>
  	<servlet-class>AjaxServlet</servlet-class>
  </servlet>
  
  <servlet-mapping>
  	<servlet-name>ajax</servlet-name>
  	<url-pattern>*.do</url-pattern>
  
  </servlet-mapping>
</web-app>

　　

posted @ 2016-10-20 18:30  DT大时代  阅读(1111)  评论(0)    收藏  举报