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Javascript 中的 任意位置显示html菜单
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脚本说明:
把如下代码加入<body>区域中
<SCRIPT language=javascript><!--
document.onclick = popUp
function popUp() {
newX = window.event.x + document.body.scrollLeft
newY = window.event.y + document.body.scrollTop
menu = document.all.itemopen
if ( menu.style.display == ""){
menu.style.display = "none" }
else {
menu.style.display = ""}
menu.style.pixelLeft = newX - 50
menu.style.pixelTop = newY - 50
}
//-->
</SCRIPT>
<!-- Start of Popup Menu -->
<table id="itemopen" class='box' style='display:none'>
<tr><td><a href='index.htm' class="cc" >弹出菜单</a></td></tr>
<tr><td><a href='time1.htm' class="cc" >本站首页</a></td></tr>
<tr><td><a href='time2.htm' class="cc">My求职信</a></td></tr>
<tr><td><a href='time3.htm' class="cc" >网上赚钱</a></td></tr>
<tr><td><a href='ye1.htm' class="cc" >免费资源</a></td></tr>
<tr><td><a href='ye2.htm' class="cc" >更多连接</a></td></tr>
<tr><td><a href='ye3.htm' class="cc" >更多连接</a></td></tr>
<tr><td><a href='zh1.htm' class="cc" >更多连接</a></td></tr>
</table>
<!-- End of Popup Menu -->
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