我又来问问题了,(自学JS没人请教也是尴尬,好像找个师傅啊),下面是问题 谢谢了
这是图片轮播代码,麻烦大家,在浏览器打开看下,这样能说清楚问题
<!DOCTYPE html>
<html>
<head lang="en">
<meta charset="UTF-8">
<title></title>
<style type="text/css">
*{margin:0; padding:0;}
li{list-style:none; background:#f60; float:left; margin-right:10px; border-radius:8px; padding:2px 6px; cursor:pointer;
text-align:center;}
ul{width:140px; position:absolute; bottom:20px; right:20px;}
.main{width:720px; height:450px; margin:0 auto; position:relative; left:0; top:100px; border:5px green solid;}
.abc{width:720px; height:106px; margin:auto; margin-top:120px; background:#ccc; text-align:center;}
.abc img{margin-right:10px; width:144px; height:90px; margin-top:8px;}
</style>
</head>
<body>
<div class="main">
<img src="1.jpg" id="tp"/>
<ul>
<li id="dianji1" onmousemove="showA(1)">1</li>
<li id="dianji2" onmousemove="showA(2)">2</li>
<li id="dianji3" onmousemove="showA(3)">3</li>
<li id="dianji4" onmousemove="showA(4)">4</li>
</ul>
</div>
<div class="abc">
<img src="1.jpg" id="pic1" onmousemove="showB(1)"/>
<img src="2.jpg" id="pic2" onmousemove="showB(2)"/>
<img src="3.jpg" id="pic3" onmousemove="showB(3)"/>
<img src="4.jpg" id="pic4" onmousemove="showB(4)"/>
</div>
<script type="text/javascript">
var gh=1;
var mc=1;
var tup=['http://a1.qpic.cn/psb?/V11aE2wE1DtywZ/I93pSorEWp54MpVhMNDEuPoY6sWU2y4sel7zcxygX0U!/b/dAMAAAAAAAAA&bo=0ALCAdACwgEBCS4!&rf=viewer_4',
'http://a2.qpic.cn/psb?/V11aE2wE1DtywZ/gYhbpYS80HxerKihb7FP8Xz6oBPopJd4Dv.Or6VNIJM!/b/dFgAAAAAAAAA&bo=0ALCAdACwgEBACc!&rf=viewer_4',
'http://a2.qpic.cn/psb?/V11aE2wE1DtywZ/kC2U6ySDDMDiJi7Dmk5TtdH7S609Ny4mhTFEO96rNjg!/b/dGQAAAAAAAAA&bo=0ALCAdACwgEBACc!&rf=viewer_4',
'http://a1.qpic.cn/psb?/V11aE2wE1DtywZ/vC9QsPEduvx6GCOIpq80bXh0t3oUAIzT2smAfm*lrwY!/b/dB4AAAAAAAAA&bo=0ALCAdACwgEBACc!&rf=viewer_4'];
document.getElementById('dianji1').style.backgroundColor='black';
document.getElementById('dianji1').style.color='white';
document.getElementById('pic1').style.border='3px #f60 solid';
function showA(a){
var dianjigh=document.getElementById('dianji'+gh);
var dianjia=document.getElementById('dianji'+a);
dianjigh.style.backgroundColor='#f60';
dianjigh.style.color='black';
document.getElementById('pic'+gh).style.border='none';
dianjia.style.backgroundColor='black';
dianjia.style.color='white';
document.getElementById('tp').src=tup[a-1];
document.getElementById('pic'+a).style.border='3px #f60 solid';
gh=a;
}
function showB(c){
var dianjimc=document.getElementById('dianji'+mc);
var dianjic=document.getElementById('dianji'+c);
dianjimc.style.backgroundColor='#f60';
dianjimc.style.color='black';
document.getElementById('pic'+mc).style.border='none';
dianjic.style.backgroundColor='black';
dianjic.style.color='white';
document.getElementById('tp').src=tup[c-1];
document.getElementById('pic'+c).style.border='3px #f60 solid';
mc=c;
}
</script>
</body>
</html>
鼠标放在 1,2,3,4上后,上面对应的图片切图,下面的小图片带边框
鼠标放在下面的小图片上,上面的大图也会切图
(只设置了onmousemove并没onmouseout)
那么问题来了
假如你鼠标放在2那,下面对应的第二个图会出现边框,然后你把鼠标移到下面第三个图上,结果下面第三个图也出现边框了,一下两个边框
同理鼠标一开始放在下面第二个图上,然后上面对应的数字2背景变黑,这个时候,鼠标移到上面数字3上面.同时3的背景色也变黑了 ,
怎么解决这个问题呢,加上onmouseout可以吗
怎样才能让第一个函数工作的时候清除第二个函数带来的效果,第二个函数工作的时候清除第一个函数带来的效果?
真心求教////
一个函数就好了吧，搞那么麻烦，mouseover的时候同时清空其他选项样式
<!DOCTYPE html>
<html>
<head lang="en">
<meta charset="UTF-8">
<title></title>
<style type="text/css">
* {
margin: 0;
padding: 0;
}
li {
list-style: none;
background: #f60;
float: left;
margin-right: 10px;
border-radius: 8px;
padding: 2px 6px;
cursor: pointer;
text-align: center;
}
ul {
width: 140px;
position: absolute;
bottom: 20px;
right: 20px;
}
.main {
width: 720px;
height: 450px;
margin: 0 auto;
position: relative;
left: 0;
top: 100px;
border: 5px green solid;
}
.abc {
width: 720px;
height: 106px;
margin: auto;
margin-top: 120px;
background: #ccc;
text-align: center;
}
.abc img {
margin-right: 10px;
width: 144px;
height: 90px;
margin-top: 8px;
}
</style>
</head>
<body>
<div class="main">
<img src="1.jpg" id="tp" />
<ul>
<li id="dianji1" onmousemove="showA(1)">1</li>
<li id="dianji2" onmousemove="showA(2)">2</li>
<li id="dianji3" onmousemove="showA(3)">3</li>
<li id="dianji4" onmousemove="showA(4)">4</li>
</ul>
</div>
<div class="abc">
<img src="1.jpg" id="pic1" onmousemove="showA(1)" />
<img src="2.jpg" id="pic2" onmousemove="showA(2)" />
<img src="3.jpg" id="pic3" onmousemove="showA(3)" />
<img src="4.jpg" id="pic4" onmousemove="showA(4)" />
</div>
<script type="text/javascript">
var now=1;
var tup = ['http://a1.qpic.cn/psb?/V11aE2wE1DtywZ/I93pSorEWp54MpVhMNDEuPoY6sWU2y4sel7zcxygX0U!/b/dAMAAAAAAAAA&bo=0ALCAdACwgEBCS4!&rf=viewer_4',
'http://a2.qpic.cn/psb?/V11aE2wE1DtywZ/gYhbpYS80HxerKihb7FP8Xz6oBPopJd4Dv.Or6VNIJM!/b/dFgAAAAAAAAA&bo=0ALCAdACwgEBACc!&rf=viewer_4',
'http://a2.qpic.cn/psb?/V11aE2wE1DtywZ/kC2U6ySDDMDiJi7Dmk5TtdH7S609Ny4mhTFEO96rNjg!/b/dGQAAAAAAAAA&bo=0ALCAdACwgEBACc!&rf=viewer_4',
'http://a1.qpic.cn/psb?/V11aE2wE1DtywZ/vC9QsPEduvx6GCOIpq80bXh0t3oUAIzT2smAfm*lrwY!/b/dB4AAAAAAAAA&bo=0ALCAdACwgEBACc!&rf=viewer_4'];
function showA(a) {
if (a == now) return;
o = document.getElementById('dianji' + now); o.style.backgroundColor = '#f60'; o.style.color = 'black';
o = document.getElementById('pic' + now); o.style.border = 'none';
o = document.getElementById('dianji' + a); o.style.backgroundColor = 'black'; o.style.color = 'white';
o = document.getElementById('pic' + a); o.style.border = '3px #f60 solid';
document.getElementById('tp').src = tup[a - 1];
now = a;
}
showA(1);//直接重函数设置默认焦点图片就好，不需要另外写代码
</script>
</body>
</html>
最简单是，获取当前选中节点的父节点，如何遍历你的四个节点全部置为none，再给当前获取到的节点为block。
通过冒泡方法来完成。冒泡是比较基础的了，如果不会，建议先去看看javascript的基础。
removeClass
可以用jQuery操作你的js,现在对原生js都不记得多少了。
参考:http://www.runoob.com/jquery/jquery-tutorial.html
<head>
<title>pic player</title>
<script type="text/javascript" src="http://img.jb51.net/jslib/jquery/jquery-1.2.6.js"></script> ......
答案就在这里：js实现图片轮播
----------------------你好，人类，我是来自CSDN星球的问答机器人小C，以上是依据我对问题的理解给出的答案，如果解决了你的问题，望采纳。