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Start your free trialDaniel Springer
Courses Plus Student 5,090 PointsI found 3 possible answers. But which one is the right one in this case?
let listItems = document.getElementsByTagName('li')
or
let listItems = document.querySelectorAll('li')
or
let listItems = document.querySelectorAll("#rainbow li");
<!DOCTYPE html>
<html>
<head>
<title>Rainbow!</title>
</head>
<body>
<ul id="rainbow">
<li>This should be red</li>
<li>This should be orange</li>
<li>This should be yellow</li>
<li>This should be green</li>
<li>This should be blue</li>
<li>This should be indigo</li>
<li>This should be violet</li>
</ul>
<script src="js/app.js"></script>
</body>
</html>
let listItems;
const colors = ["#C2272D", "#F8931F", "#FFFF01", "#009245", "#0193D9", "#0C04ED", "#612F90"];
for(var i = 0; i < colors.length; i ++) {
listItems[i].style.color = colors[i];
}
2 Answers
Matt Brock
28,330 PointsI'd use the getElementsByTagName()
method, as it is MUCH more performant in this case (see this jsperf test). querySelectorAll()
has more flexibility and can use any selector. A general rule of thumb would be:
- Use
getElementsByTagName()
when you're searching for groups of elements (<li>
,<a>
, etc.) - Use
querySelectorAll()
when you're searching for groups of complex selectors (.element .child
,#rainbow li
, etc.)
Here's a good breakdown to read through.
Daniel Springer
Courses Plus Student 5,090 PointsHi Matt, thanks for taking the time to explain.
Matt Brock
28,330 PointsYou're so welcome :)
Matt Brock
28,330 PointsMatt Brock
28,330 PointsAnother huge difference is what the two methods return when called.
getElementsByTagName()
returns a live list of HTML nodes that are updated as content within them changes.querySelectorAll()
, however, returns a non-live list of HTML nodes, or static NodeList, that is not updated with content changes. Check out the MDN Docs on NodeList to read about the two: https://developer.mozilla.org/en-US/docs/Web/API/NodeList.