Ajax很简单

来源：互联网发布：2015nba总决赛库里数据编辑：程序博客网时间：2024/04/28 19:27

最初看到ajax感觉好惊喜啊，不用刷新全部页面，就可以更改局部页面数据。真的很方法

但是以为会很难，但实际做了才知道，AJAX很简单。至少入门很简单如果你会ajax那就不要看了。更不要骂我。比我牛的人太多了

互相交流，学习。MSN:whw_dream(AT)hotmail.com

首先是jsp页面和脚本，为了方便写在一个里面

这是一个很常见的检测用户名是否存在的功能

这里用的是struts

<%@ page contentType="text/html; charset=GBK" %>
<html>
<head>
<title>
ajax
</title>
</head>
<body bgcolor="#ffffff">
<h1>
<input name="username" type="text" maxlength="20" />
<input id="chk-name-btn" type="button" value="检测帐号" onclick="testName('<%=request.getContextPath()%>')" />
<div id="view_name"></div>
</h1>
</body>
</html>
<script language="javascript">
  if (window.ActiveXObject && !window.XMLHttpRequest) {
      window.XMLHttpRequest=function() {
      return new ActiveXObject((navigator.userAgent.toLowerCase().indexOf('msie 5') != -1) ? 'Microsoft.XMLHTTP' : 'Msxml2.XMLHTTP');
   };
  }//取得XMLHttpRequest对象

function testName(path){

//path是取得系统路径
                var view_name=document.getElementById("view_name");
      var req=new XMLHttpRequest();
      if (req) {
          req.onreadystatechange=function() {
           if (req.readyState==4 && req.status==200) {//判断状态，4是已发送，200已完成
                if(req.responseText==0){
                view_name.style.color='green';
                view_name.innerHTML='该用户名可以正常使用';
                 }else if(req.responseText==1){
               view_name.style.color='red';
                   view_name.innerHTML='该用户名已经被使用';
                       }else{
                     view_name.style.color='red';
                     view_name.innerHTML='该用户名含有非法字符！';
                     }
                }
        }
                        req.open('POST', path+'/ajax.do');//struts

//req.open('POST', path+'/ajax.servlet');//servlet

                        //req.open('POST', path+'/ajax.action');//webwork

                          req.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded');
                         req.send("");//发送参数如果有参数req.send("username="+user_name);用request取得
      }
}
</script>

这个jsp页面并没有取得用户名，就是演示一下。还有<div>可以换span，具体作用问美工吧。

接受ajax请求的action。

import org.apache.struts.action.*;
import javax.servlet.http.*;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import java.io.PrintWriter;

/**
* <p>Title:AjaxAction </p>
*/

public class AjaxAction extends Action {
public ActionForward execute(ActionMapping mapping,
                               ActionForm form,
                               HttpServletRequest request,
                               HttpServletResponse response)
      throws Exception {
    PrintWriter out = response.getWriter();
    out.print(1);//ajax取得都是字符的输出。如果数据量大的话，还可以用xml来发送和接受
    return null;
}
}

struts-config.xml

<action type="test.whw.upload.AjaxAction" validate="false" scope="request" path="/ajax"/>
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如果是servlet

web.xml

<servlet>
  <servlet-name>AjaxServlet </servlet-name>
  <servlet-class>servlet.AjaxServlet </servlet-class>
</servlet>
    <servlet-mapping>
        <servlet-name>AjaxServlet </servlet-name>
        <url-pattern>/AjaxServlet .servlet</url-pattern>
    </servlet-mapping>

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 {
Logger log = Logger.getLogger(this.getClass());
public void doGet(HttpServletRequest request, HttpServletResponse response)
   throws ServletException,IOException {
  response.setContentType("text/xml; charset=GBK");
  PrintWriter out = response.getWriter();
    out.print(2);
   }
}
//Process the HTTP Post request
public void doPost(HttpServletRequest request, HttpServletResponse response)
   throws ServletException, IOException {
  doGet(request, response);
}

//Process the HTTP Put request
public void doPut(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
}

//Process the HTTP Delete request
public void doDelete(HttpServletRequest request,
HttpServletResponse response) throws ServletException, IOException {
}

//Clean up resources
public void destroy() {
}
}

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如果是webwork

xwork.xml

AjaxAction.java

import java.io.PrintWriter;

public class AjexAlbumAction extends ActionSupport implements Action{

public void ajax()throws IOException {
             PrintWriter pw = ServletActionContext.getResponse().getWriter();
            ServletActionContext.getResponse().setContentType("text/html;charset=GBK");
            pw.print(1);
            pw.close();
}

}

本例子在winXPsp2、JB9、Eclipse3，j2sdk1.4.1 、Tomcat5、Tomcat4.1都能运行