[英]How to stop Jquery dropdown from scrolling with page?
所以我有一個jQuery下拉菜單的問題,如果我向下滾動頁面,下拉菜單將顯示得更遠,而不是它們應該直接位於按鈕下方。
<!doctype html>
<html>
<head>
<title></title>
<link href="/Nexus 5/Website style.css" type="text/css" rel="stylesheet"/>
<script type="text/javascript" src="jquery.js"></script>
<script type="text/javascript">
var timeoutID;
$(function(){
$('.dropdown').mouseenter(function(){
$('.sublinks').stop(false, true).hide();
window.clearTimeout(timeoutID);
var submenu = $(this).parent().next();
submenu.css({
position:'absolute',
top: $(this).offset().top + $(this).height() + 'px',
left: $(this).offset().left + 'px',
zIndex:1000
});
submenu.stop().slideDown(300);
submenu.mouseleave(function(){
$(this).slideUp(300);
});
submenu.mouseenter(function(){
window.clearTimeout(timeoutID);
});
});
$('.dropdown').mouseleave(function(){
timeoutID = window.setTimeout(function() {$('.sublinks').stop(false, true).slideUp(300);}, 250);
});
});
</script>
</head>
<body>
<div id="header" style="position: fixed; ">
<img src="smalllogo.png" style="float:left; width:180px; height:73px;">
<div class="nav">
<ul>
<li><a href="home.html">Home</a></li>
<li><a href="Nexus 5/Nexus 5 (home).html" class="dropdown" >Nexus 5</a></li>
<li class="sublinks">
<a href="Nexus 5/Nexus 5 (info).html">Info</a>
<a href="Nexus 5/Nexus 5 (root & unlock).html">Root & Unlock</a>
<a href="Nexus 5/Nexus 5 (recovery).html">Recoveries</a>
<a href="Nexus 5/Nexus 5 (roms).html">ROMs</a>
<a href="Nexus 5/Nexus 5 (kernels).html">Kernels</a>
<a href="Nexus 5/Nexus 5 (other).html">Other</a>
</li>
<li><a href="Galaxy S4/GS4 (home).html" class="dropdown">Galaxy S4</a></li>
<li class="sublinks">
<a href="Galaxy S4/GS4 (info).html">Info</a>
<a href="Galaxy S4/GS4 (root & unlock).html">Root & Unlock</a>
<a href="Galaxy S4/GS4 (recovery).html">Recoveries</a>
<a href="Galaxy S4/GS4 (roms).html">ROMs</a>
<a href="Galaxy S4/GS4 (other).html">Other</a>
</li>
<li><a href="../about.html">About</a></li>
</ul>
</div>
</div>
然后,當滾動並向下滾動時,jquery鏈接會沿頁面向下移動。
可以從androiddevelopmentdepot.com實時查看
我不明白您為什么要計算它的最高位置。 修復頂部,可以解決您的問題嗎?
嘗試這個:
submenu.css({
position:'absolute',
top: 75,
left: $(this).offset().left + 'px',
zIndex:1000
});
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