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JavaScript jQuery Basics Working with jQuery Collections The .each() method

How do you prepend attr() for checkbox input in .each method?

With this objective exercise, I have a problem using .attr for input with checkbox type. I am able to use the .each method, however, it displays (undefined) prepending the list from the HTML.

index.html
<!DOCTYPE html>
<html lang="en">
<head>
    <title>Document</title>
    <link rel="stylesheet" href="style.css">
</head>
<body>
    <h2>My Favorite Things</h2>

    <ul class="favorite-things">
        <li>Kittens</li>
        <li>Rainbows</li>
        <li>Unicorns</li>
        <li>Sprinkles</li>
    </ul>

    <script
    src="jquery-3.2.1.min.js"></script>
    <script src="app.js"></script>
</body>
</html>
app.js
$('.favorite-things li').each(function() {
const checkbox = $(this).attr('checked');
  $(this).prepend(`(${checkbox})`);
});

2 Answers

Steven Parker
Steven Parker
231,236 Points

There isn't anything on the page that could be in a "checked" state before the code runs. Plus, the string that the instructions tell you to add can be used verbatim in your code so you don't need to select anything from the existing page code.

Hi Steven, Thanks for the reply. But unfortunately, I am not getting what you are saying. This is my understanding so far, to add the string in the .attr() as such:

$('.favorite-things li').each(function() {
const $checkbox = $(this).attr($('<input type="checkbox"/>'));
  $(this).prepend('(${checkbox})');
});
Steven Parker
Steven Parker
231,236 Points

What I was saying is that you don't need "attr" at all, and don't need to create a new variable.
You can just prepend the string directly:

  $(this).prepend('<input type="checkbox"/>');

I see, Thanks Steven.

Is that allowed because there are no arguments in the function of .each?

Steven Parker
Steven Parker
231,236 Points

The fact that "attr" isn't needed here is not related to using arguments in .each (or not).

The function in .each can take one or two arguments, but this challenge asked you to create a solution without using them.. For more details, see the jQuery documentation page on .each().