How should I select only the child <li> elements inside the parent node <nav>?

Hi Jennifer! Thanks A LOT for your help. I feel like I have commited all possible errors in this exercise.

I understood your ideas and applied them to slightly change my last attempt and it worked immediately! What I did was:

let navigationLinks = document.querySelectorAll('nav a');

I was mistakenly getting the <li> tags. Now I'm taking the <a> tags. Also, I was referencing incorrectly the tags inside the querySelectorAll element, because I was separating both tags with a comma ("nav, a") while it should just be a space ("nav a").

Then I tried to apply your other suggestion (the elegant one) to Question 2, like this:

let navigationLinks = document.querySelectorAll('nav a');
let galleryLinks = document.getElementById('gallery')[0].getElementsByTagName('a');

I'll share what happened to me because it may help other students:

Apparently that piece of code was working for Question 2, but then I received the following message: "Oops! It looks like Task 1 is no longer passing". And this message appeared even when I just used the new solution too:

let navigationLinks = document.getElementsByTagName('nav')[0].getElementsByTagName('a');
let galleryLinks = document.getElementById('gallery')[0].getElementsByTagName('a');

I was intrigued and came back to Question 1. Both ways of solving worked fine.

Then I discovered my mistake in this line:

let galleryLinks = document.getElementById('gallery')[0].getElementsByTagName('a');

Since getElementById returns only a single element, and not an array of elements, I shouldn't use the [0] position specifier. I took it out and the code then worked well finally.

Then, I moved confident to Question 3, and wrote the following mistaken code:

let navigationLinks = document.querySelectorAll('nav a');
let galleryLinks = document.getElementById('gallery').getElementsByTagName('a');
let footerImages = document.getElementsByTagName('footer').getElementsByTagName('img');

I couldn't believe that it was not working. BUMMER! In despair, I just corrected it like this:

let navigationLinks = document.querySelectorAll('nav a');
let galleryLinks = document.getElementById('gallery').getElementsByTagName('a');
let navigationLinks = document.querySelectorAll('footer img');

And it worked! I felt relieved that at least one part of my new knowledge was working. Then, suddenly I got why the former code didn't work. It was because I used the getElementsByTagName without specifying the [0] position.

Finally, I did all the three questions again, to make sure I could reproduce them with no error. Everything went fine with this code:

let navigationLinks = document.querySelectorAll('nav a');
let galleryLinks = document.getElementById('gallery').getElementsByTagName('a');
let footerImages = document.getElementsByTagName('footer')[0].getElementsByTagName('img');

This challenge was really a good challenge for me. I hope that I can retain this knowledge now and that I don't make the same mistakes again.

Once again, thank you a lot for your precious help Jennifer!

Marco Aurélio Angioluci you're a rock star! That's amazing. No, seriously, learning to debug code is somewhat of an artform and you just took a giant leap forward. I'm only glad that I was able to help