Question

某些DIV将具有名为outage的锚名称，并且那些应该将addClass outage添加到父类.modAcTitle，结果将是<div class="modAcTitle outage">This is a title.</div>我无法添加任何内容标题是独一无二的，所以这是我唯一能做的概念。

但是，我的方法非常糟糕，因为它会在每个.outage 中添加.modAcTitle，而不是在其内容中包含outage个锚点的标题。

我做错了什么？

$('a[name=outage]').parent().parent().parent().find('.modAcTitle').addClass("outage")

这是我的DIV结构和演示：https://jsfiddle.net/61za2n72/7/

<div id="ac_expand_0" class="modAc">
<div class="modAcTitle">This is a title.</div>
<div class="modAcContent" style="display: none;"><p><a name="outage"></a>This is Content and it has a unique name tag called outage</p>
</div>
<div class="modAcTitle">This is a title</div>
<div class="modAcContent" style="display: none;"><p><a name="outage"></a>This is Content and it has a unique name tag called outage</p>
</div>
<div class="modAcTitle">This is a title</div>
<div class="modAcContent" style="display: none;"><p>This is content, but it doesn't have a name tag.</p>
</div>
</div>

Answer 1

您的DOM遍历不太正确。您可以使用closest()和prev()来修复和简化它，如下所示：

$('a[name=outage]').closest('.modAcContent').prev('.modAcTitle').addClass('outage');

Updated fiddle

找到锚点并将类添加到最近的类

1 个答案: