Welcome to the Treehouse Community

Want to collaborate on code errors? Have bugs you need feedback on? Looking for an extra set of eyes on your latest project? Get support with fellow developers, designers, and programmers of all backgrounds and skill levels here with the Treehouse Community! While you're at it, check out some resources Treehouse students have shared here.

Looking to learn something new?

Treehouse offers a seven day free trial for new students. Get access to thousands of hours of content and join thousands of Treehouse students and alumni in the community today.

Start your free trial

JavaScript jQuery Basics Working with jQuery Collections Working with jQuery-Specific Selectors

Josh Hamilton
seal-mask
.a{fill-rule:evenodd;}techdegree
Josh Hamilton
Front End Web Development Techdegree Student 5,201 Points
Posted by Josh Hamilton
.a{fill-rule:evenodd;}techdegree
Josh Hamilton
Front End Web Development Techdegree Student 5,201 Points

This code comes back as an error and I've switched it around as many times as I could think of. Any suggestions?

I've changed the (const) to (var); and i've also removed the dollar sign in front of the variable name to see if that would work and it didn't. Besides that I can't think of any other reason why it wouldn't be accepted.

index.html 
<!DOCTYPE html>
<html lang="en">
<head>
    <title>Document</title>
    <link rel="stylesheet" href="style.css" type="text/css" media="screen" title="no title" charset="utf-8">
</head>
<body>
    <h2>Star Trek Characters</h2>

    <ul class="character-list">
        <li>Captain Jean Luc Picard</li>
        <li>Data</li>
        <li>Warf</li>
        <li>Dr. Crusher</li>
    </ul>

    <div>I am supposed to stay hidden!</div>    

    <script
    src="jquery-3.2.1.min.js"></script>
    <script src="app.js"></script>
</body>
</html>
app.js 
const $hiddenTags = $('li:hidden');
$hiddenTags.show();

1 Answer

Steven Parker
Steven Parker
231,236 Points
Steven Parker
Steven Parker
231,236 Points

Your code is actually OK, but the challenge apparently expects it to be done without using an intermediate variable.

Attach the "show" method directly to the jQuery selector function (on the same line) and you should pass the challenge.

If you feel this is a bug in the challenge, you may want to report it to the Support staff.

Steven Parker
Steven Parker
231,236 Points
Steven Parker
Steven Parker
231,236 Points

Use just the right-hand part. You don't need to create a variable or assign to it.

Steven Parker
Steven Parker
231,236 Points
Steven Parker
Steven Parker
231,236 Points

Glad to help. You can mark a question solved by choosing a "best answer".

And happy coding!