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          ---------------------------------AJAX.html-------------------------------------------
          <html>
          ?<head>
          ??<title>AJAX.html</title>
          ?</head>
          ?<script type="text/javascript">
          ?/*通過異步傳輸XMLHTTP發送參數到ajaxServlet,返回符合條件的XML文檔*/
          ?function getResult(){
          ??var url = "ajaxServlet?action=send";
          ??if (window.XMLHttpRequest){
          ???req = new XMLHttpRequest();
          ??}else if (window.ActiveXObject)
          ??{
          ???req = new ActiveXObject("Microsoft.XMLHTTP");
          ??}
          ??if(req)
          ??{
          ???req.open("GET",url, true);
          ???req.onreadystatechange = complete;
          ???req.send(null);
          ??}
          ?}
          ?/*分析返回的XML文檔*/
          ?function complete(){
          ??if (req.readyState == 4)
          ??{
          ???if (req.status == 200){
          ????var type = req.responseXML.getElementsByTagName("type_name");
          ????var str=new Array();
          ????for(var i=0;i<type.length;i++){
          ?????str[i]=type[i].firstChild.data;
          ?????document.all['td'].innerHTML+=str[i]+"<BR>";
          ????}
          ???}
          ??}
          ?}
          </script>
          ?<body onLoad="getResult()">
          ??<table width="80%" border="0" cellspacing="0" cellpadding="0">
          ???<tr>
          ????<td id="td">
          ????</td>
          ???</tr>
          ??</table>
          ?</body>
          </html>
          ---------------------------------ajaxServlet.java-----------------------------------------------------
          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 String CONTENT_TYPE = "text/xml; charset=gb2312";
          ?public void init() throws ServletException{}
          ?public void doGet(HttpServletRequest request,HttpServletResponse response) throws ServletException, IOException
          ?{
          ??response.setContentType(CONTENT_TYPE);
          ??PrintWriter out = response.getWriter();
          ??String action = request.getParameter("action");
          ??if (("send").equals(action))
          ??{
          ???StringBuffer sb = new StringBuffer("<type>");
          ???sb.append("<type_name>AA</type_name>");
          ???sb.append("<type_name>BB</type_name>");
          ???sb.append("<type_name>CC</type_name>");
          ???sb.append("<type_name>DD</type_name>");
          ???sb.append("</type>");
          ???out.write(sb.toString());
          ???out.close();
          ??}
          ?}
          }
          --------------------------------web.xml------------------------------------------------------
          <?xml version = '1.0' encoding = 'UTF-8'?>
          <!DOCTYPE web-app PUBLIC "-//Sun Microsystems, Inc.//DTD Web Application 2.3//EN" "<web-app>
          ?<description>Empty web.xml file for Web Application</description>
          ?<servlet>
          ??<servlet-name>ajax</servlet-name>
          ??<servlet-class>ajaxServlet</servlet-class>
          ?</servlet>
          ??? <servlet-mapping>
          ??????? <servlet-name>ajax</servlet-name>
          ??????? <url-pattern>/ajaxServlet</url-pattern>
          ??? </servlet-mapping>
          </web-app>


          FeedBack:
          # re: ajax簡單實例(js內產生xmlHttpRequest,它和servlet交互返回xml顯示在頁面)[未登錄]
          2012-08-15 15:03 | 菜鳥
          很好,如果用struts框架,怎么寫struts.xml文件找到servlet呢  回復  更多評論
            
          # re: ajax簡單實例(js內產生xmlHttpRequest,它和servlet交互返回xml顯示在頁面)[未登錄]
          2015-01-13 11:53 | p
          pp  回復  更多評論
            

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