javascript实现跳转菜单的具体方法
人气:0传统
这里要做的是，省略Go There按钮，选择菜单项后，直接跳转。
Html代码
复制代码 代码如下:
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>转移菜单</title>
<script type="text/javascript" src="转移菜单.js"></script>
</head>
<body>
<form>
<select id="newLocation">
<option selected="selected" value="">Select a topic</option>
<option value="topic1.html">topic1</option>
<option value="topic2.html">topic2</option>
<option value="topic3.html">topic3</option>
<option value="topic4.html">topic4</option>
<option value="topic5.html">topic5</option>
</select>
<!--当浏览器不支持javascript或者禁用脚本运行时被调用-->
<noscript>
<input type="submit" value="Go There!"/>
</noscript>
</form>
</body>
</html>
Javascript脚本
复制代码 代码如下:
window.onload=initForm;
//防止页面缓存，无法触发onload
window.onunload=function(){}
function initForm(){
document.getElementById("newLocation").selectIndex=0;
document.getElementById("newLocation").onchange=jumpPage;
}
function jumpPage(){
var newLoc=document.getElementById("newLocation");
var newPage=newLoc.options[newLoc.selectedIndex].value;
if (newPage!=""){
window.location=newPage;
}
}
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