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点了超链接后,从底下跑到顶上_html/css_WEB-ITnose

WBOY
WBOYOriginal
2016-06-21 09:35:591323browse



这是代码:用的#
".$r['number']."
二个层如下:
<div id="light" class="layContent"> 	<h2><a href="javascript:viod(0);" onclick="showOnOff('light','fade',0)">close</a></h2>	<ul>    </ul></div> <div id="fade" class="layWhole"></div>


.layWhole{	display: none;	position:absolute;	top:0%;	left:0%;	width:100%;height:100%;	background-color:black;	z-index:1001;	-moz-opacity:0.7;	opacity:.70;/*不透明*/	filter:alpha(opacity=70);}.layContent{	display:none;	position:absolute;	top:25%;	left:25%;	width:50%;	height:50%;	padding:16px;	border:5px solid orange;	background-color:white;	z-index:1002;	overflow:auto;}


回复讨论(解决方案)

".$r['number']."

".$r['number']."

".$r['number']."


把a标签里href值修改为“javascirpt:void(0);”这样在点击时,页面就不会滚动了

把a标签里href值修改为“javascirpt:void(0);”这样在点击时,页面就不会滚动了

不行的,你自己试试

应该是可以的。
".$r['number']."

".$r['number']."

".$r['number']."
多谢好汉,相助!问题仍未解决。我真的服了。不知道如何办?

首先 
XXX
这样 可以避免 页内定位

然后 你只是要执行一个点击事件 所以不用a也可以啊

XXX

加个样式就可以了
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