我有一个简单的组织列表,其中唯一的结构差异是某些父div上的类“成员”。我想根据父类的类删除一些子元素。
<div class="views-row views-row-1 views-row-odd member">
<div class="views-field views-field-title">
<span class="field-content"><a href="[Link 1 goes here]">[Name 1 goes here]</a></span>
</div>
<div class="views-field views-field-field-phone field-content">
[Phone 1 goes here]
</div>
<div class="views-field views-field-field-email field-content">
[Email 1 goes here]
</div>
<div class="views-field views-field-field-description field-content">
[Description 1 goes here]
</div>
</div>
<div class="views-row views-row-2 views-row-even">
<div class="views-field views-field-title">
<span class="field-content"><a href="[Link 2 goes here]">[Name 2 goes here]</a></span>
</div>
<div class="views-field views-field-field-phone field-content">
[Phone 2 goes here]
</div>
<div class="views-field views-field-field-email field-content">
[Email 2 goes here]
</div>
<div class=
"views-field views-field-field-description field-content">
[Description 2 goes here]
</div>
</div>
因为第1组包含类“成员”,所以我希望它保持不变。但是,由于第2组不是会员,我想删除带有.views-field-field-phone和.views-field-field-email类的div。
我相信我需要使用.not创建一个函数。这样:
$(".views-row").not(".member").css("border", "5px solid red");
会正确勾勒出正确的父母,但我很难针对孩子。
有任何见解。
谢谢。
答案 0 :(得分:1)
jQuery find应该有所帮助:
具体是:
$(".views-row").not(".member").find(".views-field-field-phone").remove();
$(".views-row").not(".member").find(".views-field-field-email").remove();
答案 1 :(得分:0)
我想这应该做你正在寻找的事情:
$("div.views-row").not(".member").find(".views-field-field-phone, .views-field-field-email").css("color", "red");
编辑:如果要删除,只需在末尾调用remove()方法而不是css
$("div.views-row").not(".member").find(".views-field-field-phone, .views-field-field-email").remove();
看看这个jsBin: