Ajax只能导航,展示链接内容

2009-03-05 12:08:31 作者:Michael | 【

您可以使用此脚本代替frames的工作,当用户点击指定链接是,链接地址的内容将在页面内显示,网页也不会刷新,如果链接地址不存在,则会显示一个友好的提示信息

查看演示页面 View Demo 查看全部代码 View Code

创建一个JS文件,ajax_navagation.js
粘贴一下代码到js文件

var please_wait = null;

function open_url(url, target) {
 	if ( ! document.getElementById) {
  		return false;
 	}

 	if (please_wait != null) {
  		document.getElementById(target).innerHTML = please_wait;
 	}

 	if (window.ActiveXObject) {
  		link = new ActiveXObject("Microsoft.XMLHTTP");
 	} else if (window.XMLHttpRequest) {
  		link = new XMLHttpRequest();
 	}

 	if (link == undefined) {
  		return false;
 	}
 	link.onreadystatechange = function() { response(url, target); }
 	link.open("GET", url, true);
 	link.send(null);
}

function response(url, target) {
 	if (link.readyState == 4) {
	 	document.getElementById(target).innerHTML = (link.status == 200) ? link.responseText : "Ooops!! A broken link! Please contact the webmaster of this website ASAP and give him the fallowing errorcode: " + link.status;
	}
}

function set_loading_message(msg) {
 	please_wait = msg;
}



在网页<head>区添加以下代码
<script type="text/javascript" src="ajax_navagation.js"></script>


在网页<body>区添加以下代码

<table>
<tr>
<td valign=top width=150>
<H5>My Navagation links</H5>
<a href="javascript:void(0)" onclick="open_url('page-1.html','my_site_content');">Go to page 1</a><br>
<a href="javascript:void(0)" onclick="open_url('page-2.html','my_site_content');">Go to page 2</a><br>
<a href="javascript:void(0)" onclick="open_url('page-3.html','my_site_content');">Go to page 3</a><br>
<a href="javascript:void(0)" onclick="open_url('page-4.html','my_site_content');">Go to page 4</a><br>
<a href="javascript:void(0)" onclick="open_url('page-5.html','my_site_content');">Go to Page 5</a><br>
</td>
<td valign=top>
<div id="my_site_content">
</div>
</td>
</tr>
</table>


网页调用了page-1.html,Page-2.html等页面的内容显示,如果页面不存在则返回一个友好信息

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