如何在 javascript 中获取特定 HTML 标签的所有元素？
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How do I get all elements of a particular HTML tag in javascript?
提问by Tom
I need to hide all elements of type 'section' in my document apart from one with a particular ID.
除了具有特定 ID 的元素之外，我需要隐藏文档中所有类型为“section”的元素。
In jquery this would be easy
在 jquery 中，这很容易
$("section").hide();
$("section#myId").show();
How would I do this without jquery??
如果没有 jquery，我将如何做到这一点？
(I need it to happen as soon as the page loads and to not be noticable). I also need it to work cross browser.
（我需要它在页面加载后立即发生并且不会引起注意）。我还需要它跨浏览器工作。
Thanks.
谢谢。
回答by Jacob Relkin
DOMElement.getElementsByTagNameis your friend:
DOMElement.getElementsByTagName是你的朋友：
var sections = document.getElementsByTagName('section');
var mySection = null;
for(var i = 0; i < sections.length; ++i) {
if(sections[i].id === "myId") {
mySection = sections[i];
mySection.style.display = "block";
break;
}
sections[i].style.display = "none";
}
回答by HBP
Place the following immediately before the </body> in your HTML
将以下内容紧邻 HTML 中的 </body> 之前
<script>
(function () {
for(var els = document.getElementsByTagName ('section'), i = els.length; i--;)
els[i].id !== "myId" && (els[i].style.display = "none");
}) ();
</script>
or in "modern" (HTML5) browsers :
或在“现代”（HTML5）浏览器中：
<script>
[].forEach.call (document.querySelectorAll ('section'),
function (el) { el.id !== "myId" && (el.style.display = "none"); })
</script>
