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JavaScript JavaScript and the DOM (Retiring) Getting a Handle on the DOM Selecting Multiple Elements

Daniel Springer
PLUS
Daniel Springer
Courses Plus Student 5,090 Points

I 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");
index.html
<!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>
js/app.js
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

I'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.

Another 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.

You're so welcome :)