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 (2014) Creating a Simple Lightbox Perform: Part 2

Rhett Herring
PLUS
Rhett Herring
Courses Plus Student 11,023 Points
Posted by Rhett Herring
Rhett Herring
Courses Plus Student 11,023 Points

$overlay.show() does not work for me.. I had to use $("#overlay").show();

This would not work for me:

var $overlay = '<div id="overlay"></div>';
$("body").append($overlay);

$("#imageGallery a").click(function(event) {
event.preventDefault();
var href = $(this).attr("href");
console.log(href);
$overlay.show();
});

I had to change $overlay.show() to $("#overlay").show(); in order to get this to work.

My CSS:

    #overlay {
        background: rgba(0,0,0, 0.7);
        width: 100%;
        height: 100%;
        position: absolute;
        top: 0;
        left: 0;
        display: none;
    }

2 Answers

Steven Parker
Steven Parker
229,744 Points
Steven Parker
Steven Parker
229,744 Points

:point_right: It looks like your top line is missing the part that creates a JQuery object from the string.

You probably meant to write:

var $overlay = $('<div id="overlay"></div>');

I'm surprised that $("#overlay").show(); worked - it could be that you also forgot to remove the overlay div from the HTML.

Ashenafi Ashebo
Ashenafi Ashebo
15,021 Points
Ashenafi Ashebo
Ashenafi Ashebo
15,021 Points

var $overlay = $('<div id="overlay"></div>'); use this on the top your jquery but not this one you used in your code " var $overlay = '<div id="overlay"></div>';"