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JavaScript JavaScript and the DOM (Retiring) Traversing the DOM Child Traversal

Victor Kandalov
Victor Kandalov
21,582 Points

bug?

JavaScript and the DOM -> Traversing the DOM -> Child Traversal

Here's the code I write

const section = document.querySelector('section'); let paragraphs = section.children; for ( i=0; i<paragraphs.length; i++ ) { paragraphs[i].style.color = 'blue'; }

It works fine in my code editor, but in your it says "Oops! It looks like Task 1 is no longer passing."

Please let me know if it's a bug on your side. Thanks.

app.js
const section = document.querySelector('section');
let paragraphs = section.children;
for ( i = 0; i < paragraphs.length; i++ ) {
    section.children[i].style.color = 'blue';
}
index.html
<!DOCTYPE html>
<html>
    <head>
        <title>Child Traversal</title>
    </head>
    <body>
        <section>
            <p>This is the first paragraph</p>
            <p>This is a slightly longer, second paragraph</p>
            <p>Shorter, last paragraph</p>
        </section>
        <footer>
            <p>&copy; 2016</p> 
        </footer>
        <script src="app.js"></script>
    </body>
</html>

2 Answers

Hi Victor,

const section = document.querySelector('section');
let paragraphs = section.children;
for ( i = 0; i < paragraphs.length; i++ ) {  // your (i) variable needs to set like this (let i = 0;)
    section.children[i].style.color = 'blue'; // this line it should be like " paragraphs[i].style.color = 'blue'; "
}
const section = document.querySelector('section');
let paragraphs = section.children;
for(let i = 0; i < paragraphs.length; i++) {
  paragraphs[i].style.color = 'blue';
}
Victor Kandalov
Victor Kandalov
21,582 Points

Thanks, you are right that it should be "let i = 0" instead of "i = 0". Concerning second line, I believe it doesn't matter, since paragraphs = section.children, so both section.children[i] and paragraphs[i] should work identically. Thanks for your help, I managed to pass the task after all, still I beleive there was a bug not related to my code itself.