What does $($(this)) mean?

Question

I saw some code around the web that uses the following statement

if ($($(this)).hasClass("footer_default")) {       $('#abc')         .appendTo($(this))         .toolbar({position: "fixed"});     } 

What is the use of $($(this)) and why is that necessary here?

Answer 1

Yes, $($(this)) is the same as $(this), the jQuery() or $() function is wonderfully idempotent. There is no reason for that particular construction (double wrapping of this), however, something I use as a shortcut for grabbing the first element only from a group, which involves similar double wrapping, is

$($('selector')[0])

Which amounts to, grab every element that matches selector, (which returns a jQuery object), then use [0] to grab the first one on the list (which returns a DOM object), then wrap it in $() again to turn it back into a jQuery object, which this time only contains a single element instead of a collection. It is roughly equivalent to

document.querySelectorAll('selector')[0];, which is pretty much document.querySelector('selector');

Answer 2

You can wrap $ as many times as you want, it won't change anything.

If foo is a DOM element, $(foo) will return the corresponding jQuery object.

If foo is a jQuery object, $(foo) will return the same object.

That's why $($(this)) will return exactly the same as $(this).