Shouldn't $('.external a') and $('a.external') evaluate the same? Using the first class selector fails the challenge.

Daniel,

It's easy to get confused with these selectors, as they do look similar. The key to understanding their differences is remembering that order matters! For consistency and simplicity, JavaScript (and jQuery) use the same selector formatting as CSS.

".example p" says "give me all of the 'p' elements contained within the '.example' element." Due to the order they are in, we are looking first at the ".example" element, then for all "p" elements contained inside it.

"p.example", on the other hand, says "look at all 'p' elements that have the class '.example'."

This is the same way that you would target things in your CSS. If you want to target all paragraph elements, you would write:

(in CSS)

/* targets all p elements */
p {
}

(in jQuery)

$( 'p' );

If, instead, you only wanted to target the paragraph elements that contained the class "special", you would write:

(in CSS)

/* targets only the p elements that have the class "special" */
p.special {
}

(in jQuery)

$( 'p.special' );

If, alternatively, you had a particular container on your page, let's say it has the class name "quote-of-the-day", and you wanted to style the paragraph tags inside that container to stand out from the rest of the paragraphs on your page, you would write:

(in CSS)

/* this first targets the "quote-of-the-day" container, then more specifically, it targets all the p tags contained therein */
.quote-of-the-day p {
}

(in jQuery)

$( '.quote-of-the-day p' );

Notice that the CSS and jQuery selectors are all the exact same. I hope this helps to clarify!

Erik