三个div(left,mid,right)是用float属性并列显示在界面的。
我复制了class=content这个节，如果我用eq（）指定改或删除我复制的节点的类名（eg.left,mid,right），竟然原来的节点的类名也会改变？要怎么让原来的不变呢？不知道是不是float这个属性造成的？请指教
我是这样操作节点的
$('.all').append( $('.content:eq(-1)').clone(true));
$(".all>.content:eq(-2)").children('.left').removeClass("left");
$(".all>.content:eq(-2)").children('.right').removeClass("right");
$(".all>.content:eq(-2)").children('.mid').removeClass("mid");
$(".right").html("");
$(".left").html("");
$(".mid").html("");
我最终要实现的效果是，服务器获取的数据（each（）遍历获得的）要写入到不同的div里面，所以想着改变我不需要写入的div的属性，就不会写入了，不知道思路对了没？
回复讨论(解决方案)
那是你找到的是所有的，可能并没有指向它或它的结点、你的方法并没有错
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"><html> <head> <title>MyHtml.html</title> <meta http-equiv="keywords" content="keyword1,keyword2,keyword3"> <meta http-equiv="description" content="this is my page"> <meta http-equiv="content-type" content="text/html; charset=UTF-8"> </head></head><body> <div class="all" id="scroll"> <DIV class="content"> <div class="left">1</div> <div class="mid">2</div> <div class="right">3</div> </DIV> </div><script src="http://libs.baidu.com/jquery/1.11.1/jquery.min.js"></script> <script> $(function() { //console.log('-------------------------------'); $('.all').append($('.content:eq(0)').clone(true)); $(".all>.content:eq(1)").children('.left').removeClass("left"); $(".all>.content:eq(1)").children('.right').removeClass("right"); $(".all>.content:eq(1)").children('.mid').removeClass("mid"); $(".right").html("x");// 此时第二个div的class 删除，没有影响下面的数据。1、2、3并没有变 $(".left").html("y"); $(".mid").html("z"); }); </script> </body> </html> 思路不是很明确，你可以换另一种逻辑来处理
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