ajax+jsp+servlet 实现动态显示系统当前时间
1. showCurrentTime.jsp<%@ page language="java" contentType="text/html; charset=gb2312" pageEncoding="gb2312"%><!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd"><html><head><meta http-equiv="Content-Type" content="text/html; charset=gb2312"><title>动态显示系统当前时间</title></head><% String path = request.getContextPath();String basePath = request.getScheme()+"://"+request.getServerName()+":"+request.getServerPort()+path+"/";%><script>var xmlHttp;function createXMLHttpRequest(){if(window.ActiveXObject){xmlHttp = new ActiveXObject("Microsoft.XMLHttp");}else if(window.XMLHttpRequest){xmlHttp = new XMLHttpRequest();}}function exec(){createXMLHttpRequest();xmlHttp.open("post","<%=basePath %>"+"timeUpdate",true);xmlHttp.onReadyStateChange = handleFun;xmlHttp.send(null);}function handleFun(){if(xmlHttp.readyState == 4){if(xmlHttp.status == 200){document.getElementById("div").innerText = xmlHttp.responseTEXT;}}}setInterval("exec()",1000);</script><body>你好！TimeUpdateServlet路径:<%=basePath %><div id="div"></div></body></html>
2. TimeUpdateServlet.java
package com.ibm.user.web;import java.io.IOException;import java.io.PrintWriter;import java.text.DateFormat;import java.text.SimpleDateFormat;import java.util.Date;import javax.servlet.ServletException;import javax.servlet.http.HttpServlet;import javax.servlet.http.HttpServletRequest;import javax.servlet.http.HttpServletResponse;public class TimeUpdateServlet extends HttpServlet {private static final long serialVersionUID = 1L;protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {request.setCharacterEncoding("gb2312");response.setContentType("text/html;charset=gb2312");Date date = new Date();DateFormat dformater = new SimpleDateFormat("yyyy-MM-dd hh:mm:ss");String currentTime = dformater.format(date);PrintWriter out = response.getWriter();out.println(currentTime);out.flush();out.close();}protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {doPost(request,response);}}
3.web.xml
<?xml version="1.0" encoding="UTF-8"?><web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://java.sun.com/xml/ns/javaee" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" id="WebApp_ID" version="2.5"><display-name>xg3</display-name> <servlet> <description></description> <display-name>TimeUpdateServlet</display-name> <servlet-name>TimeUpdateServlet</servlet-name> <servlet-class>com.ibm.user.web.TimeUpdateServlet</servlet-class></servlet><servlet-mapping> <servlet-name>TimeUpdateServlet</servlet-name> <url-pattern>/timeUpdate</url-pattern></servlet-mapping></web-app>
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