我对JS很不错。 我需要使用html代码选择data-parent =“16”的元素:
<div class="filter-options" id="able-filter">
<div class="Xhaut" id="Xhaut">
<a class=" active" data-dimension="region" data-filter="all" href="javascript:void(0)">All</a>
<a data-term_id="18" data-parent="17" data-filter="villebelge1" href="javascript:void(0)" class="a" data-dimension="region">villeBelge1</a>
<a data-term_id="19" data-parent="17" data-filter="villebelge2" href="javascript:void(0)" class="a" data-dimension="region">villeBelge2</a>
<a data-term_id="20" data-parent="17" data-filter="villebelge3" href="javascript:void(0)" class="b" data-dimension="region">villeBelge3</a>
<a data-term_id="21" data-parent="16" data-filter="villenord1" href="javascript:void(0)" class="b" data-dimension="region">villenord1</a>
<a data-term_id="22" data-parent="16" data-filter="villenord2" href="javascript:void(0)" class="c" data-dimension="region">villenord2</a>
<a data-term_id="23" data-parent="15" data-filter="villesud1" href="javascript:void(0)" class="c" data-dimension="region">villeSud1</a>
<a data-term_id="24" data-parent="15" data-filter="villesud2" href="javascript:void(0)" class="d" data-dimension="region">villeSud2</a>
<a data-term_id="25" data-parent="15" data-filter="villesud3" href="javascript:void(0)" class="d" data-dimension="region">villeSud3</a>
</div>
</div>
我试过这种事:
var Xhaut = $('#Xhaut').find('a');
window.console && console.log(Xhaut);
window.console && console.log(Xhaut.siblings(".d")); // thi it ok I have villesud2 and villesud3, it's ok because I use a class selector
// But how can I select all element with data-parent="16" (not class="d", but data-parent="16")
我做了一个小提琴: http://jsfiddle.net/bakalegum/2DGHD/
很多。
编辑:为你的答案而战!
答案 0 :(得分:2)
你可以这样做:
var Xhaut = $('#Xhaut').find('[data-parent=16]');
答案 1 :(得分:0)
没有jQuery:
var Xhaut = document.querySelectorAll('[data-parent="16"]');
https://developer.mozilla.org/docs/DOM/Document.querySelectorAll
答案 2 :(得分:0)
只要扩展tymeJV的内容,你就可以将其缩短一点:
var Xhaut = $('[data-parent=16]','#Xhaut');