Question

Yeah, the way I was doing is

   A.add(B).add(C).add(D).show()

While A,B,C,D are jQuery objects. I wonder if there’s such a simpler way to do this out there?
I’ve tried all the following approaches, but no results:

$(A,B,C,D).show()
A.add(B,C,D).show()

All suggestions are welcome!

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Addition to clarify the question:

The part “.show()” is just for demonstration. I just wanted to know how could I create a set of JQuery object like $(‘p’) create a set of p tag.

In my real case, I used

$([A,B,C,D]).each(fn)

instead of .show() (And I wonder why this worked?)
It’s obviously that

$([A,B,C,D]).each(fn)
$('p').each(fn)

both work. But

$([A,B,C,D]).show()  //--doesn't work
$('p').show()        //--works

Just the second line works. Does anyone know the diffrence between them ?
(I just thought they’re the same, then made a bit of tangle in my question)

Answer 1

$.each([A,B,C,D], function(){ 
    $(this).css('background','red'); 
})

Instead of using a selector this solution uses a jQuery method called $.each which takes an array and iterates over it. The array passed is a set of jQuery objects. $(this) referes to the jQuery objects being iterated on each time.

demo: http://jsfiddle.net/SCjMc/1/

---

Other facts about how $() works:

$(element) is an shortcut for jQuery(element). The jQuery() method accepts a different set of parameters:

The description for each type of parameter can be found in this link.

One of the type of parameters is “elementArray”. The description for this is:

elementArray An array containing a set of DOM elements to wrap in a jQuery object.

The catch is that when you use jQuery to select an element this returns a jQuery object and not a DOM Element directly. Therefore this will not return any elements:

var element1 = $("selector1");
var element2 = $("selector2");
$([element1,element2]) // will not return any elements

To return a DOM Element instead of a jQuery object you have to access the property in the position of 0 of the jQuery object. As follows: $("element")[0]. And that is why this will work:

var element1 = $("selector1")[0]; //accessing dom element 
var element2 = $("selector2")[0]; //from jQuery object
$([element1,element2])