在data-Attribute javascript上使用querySelectorAll（无jQuery）

Question

我知道如何使用querySelectorAll选择特定的类。 但是，我该如何选择具有特定数据属性的DOM对象。

例如：

 <div class="person">
    <div class="detail" data-field="name">Tim</div>
    <div class="detail" data-field="age">24</div>
    <div class="detail" data-field="hair">black</div>
</div>
<div class="person">
    <div class="detail" data-field="name">Tim</div>
    <div class="detail" data-field="age">34</div>
    <div class="detail" data-field="name">red</div>
</div>
<div class="person">
    <div class="detail" data-field="name">David</div>
    <div class="detail" data-field="age">56</div>
    <div class="detail" data-field="name">brown</div>
</div>

如果我要选择具有特定类的DOM，例如“详细信息”

 document.querySelectorAll('.detail')

我的问题是，如何为所有带有“ data-field = name”的域名选择？

Answer 1

您要使用属性选择器

 // every element with a data-field attribute var dataFieldElements = document.querySelectorAll('[data-field]'); console.log(dataFieldElements); // only those elements that have their data-field attribute equal to name var dataFieldNameElements = document.querySelectorAll('[data-field=name]'); console.log(dataFieldNameElements); 

 <div class="person"> <div class="detail" data-field="name">Tim</div> <div class="detail" data-field="age">24</div> <div class="detail" data-field="hair">black</div> </div> <div class="person"> <div class="detail" data-field="name">Tim</div> <div class="detail" data-field="age">34</div> <div class="detail" data-field="name">red</div> </div> <div class="person"> <div class="detail" data-field="name">David</div> <div class="detail" data-field="age">56</div> <div class="detail" data-field="name">brown</div> </div> 

Answer 2

使用属性选择器

[attr = value]

表示一个属性名称为attr且元素的值恰好是“值”的元素。

document.querySelectorAll('.detail[data-field="name"]')