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JavaScript DOM Scripting By Example Editing and Filtering Names Filter Invitees Who Have Not Responded

Posted by Daniel Guy
Daniel Guy
15,305 Points

I thought that it may be quite excessive to loop through ul.children twice so wanted to come up with a dryer solution...

filterCheckbox.addEventListener('change', (e) => { const isChecked = e.target.checked; const lis = ul.children; for (let i=0; i < lis.length; i += 1) { let li = lis[i]; if (isChecked) { if (li.className === 'responded') { li.style.display = ''; } else { li.style.display = 'none'; } } else { li.style.display = ''; } } });

This seems to work but I was wondering if there was a reason it wasn't done this way on the tutorial?

2 Answers

Victor Mercier
Victor Mercier
14,667 Points

Hi, in programming there are often many approach to a problem, so sometimes you might end with a better solution than the teacher!

Erland Van Olmen
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Erland Van Olmen
Full Stack JavaScript Techdegree Graduate 17,825 Points
Erland Van Olmen
.a{fill-rule:evenodd;}techdegree seal-36
Erland Van Olmen
Full Stack JavaScript Techdegree Graduate 17,825 Points

In theory, at least in a compiled language (which Javascript isn't), your solution would be slower. Not that you would be able to measure the difference ;) In traditional programming languages, one will always try to remove unnecessary checks from loops, especially if these loops are nested. It is best practice if you will, the instructor probably does it out of habit.