javascript 通过标签名和类名获取元素
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Get element by tag name and class name
提问by Leke
(in vanilla JavaScript) I was wondering if the was an easy way to do something like
（在 vanilla JavaScript 中）我想知道这是否是一种简单的方法来做类似的事情
x = document.getElementsByTagName('span') && getElementsByClassName('null');
To return all 'span' elements that have the class name 'null'?
返回所有类名为“null”的“span”元素？
I thought it might have been something like:
我认为它可能是这样的：
x = document.getElementsByTagName('span');
x = x.getElementsByClassName('null');
// or
x = document.getElementsByTagName('span').getElementsByClassName('null');
But that didn't seem to work out.
但这似乎并没有奏效。
Is this possible or will I have to iterate through x popping anything that returns false for .class='null'?
这是可能的还是我必须遍历 x 弹出任何为 .class='null' 返回 false 的东西？
Thanks.
谢谢。
回答by SLaks
The DOM does not provide any APIs for filtering NodeLists.
DOM 不提供任何用于过滤 NodeList 的 API。
Instead, you can use CSS selectors:
相反，您可以使用 CSS 选择器：
var x = document.querySelectorAll('span.null');
