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

Lucas Huerta-Murillo Grau
PLUS
Lucas Huerta-Murillo Grau
Courses Plus Student 1,365 Points
Posted by Lucas Huerta-Murillo Grau
Lucas Huerta-Murillo Grau
Courses Plus Student 1,365 Points

I can't find the bug.

I can't see what I am doing wrong.

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 = document.querySelectorAll('#rainbow');
const colors = ["#C2272D", "#F8931F", "#FFFF01", "#009245", "#0193D9", "#0C04ED", "#612F90"];

for(var i = 0; i < colors.length; i ++) {
  listItems[i].style.color = colors[i];    
}
Reid Everett
Reid Everett
14,009 Points
Reid Everett
Reid Everett
14,009 Points

You need to select all of the "list items" from the ID of the element you're selecting. Line 1 is missing something and get rid of the space from "i ++"

Steven Parker
Steven Parker
231,269 Points
Steven Parker
Steven Parker
231,269 Points

Actually, while it may look odd, the space before the post-increment operator doesn't cause a problem, and changing it is not part of this challenge.

1 Answer

Steven Parker
Steven Parker
231,269 Points
Steven Parker
Steven Parker
231,269 Points

You would not normally use querySelectorAll with a single ID.

Since an ID must be unique, you would most likely want to get a single element instead of a collection.

But in this case you really do want a collection, and specifically list items ("li"). So you might want to use a descendant selector combining that type with your list ID. Or you could get the list as a single element (with a different method) and then take the children property of that.