Passing multiple jQuery objects as an array

Question

I'm wondering if it's possible to do something like this in jQuery:

$( [ $selector1, $selector2, document.getElemetByID( 'test' ) ] )

EDIT: I'm lazy. $selector1 and $selector2 are already jQuery objects. I am trying to avoid using $.each.

Here's a more in-depth example of what I'm trying to do:

var $lastLink = $( '.links' ).filter( ':last' );
var $nextThumbnail = $( '.thumbnail.current' ).next();

Let's say I want to add a class of active to those things and I would like to keep it as simple as possible - therefore I am looking for something with similar syntax to what I posted before.

$( [ $lastLink, $nextThumbnail ] ).addClass( 'active' );

What is it that you are trying to do with that? What are you hoping for as the result? — talemyn
– talemyn, Commented Mar 6, 2015 at 22:11
$("selector1", "selector2", document.getElementById("test")) comma delimit for multiple selectors — Sterling Archer
– Sterling Archer, Commented Mar 6, 2015 at 22:12
@SterlingArcher - Maybe . .. or he could be looking for $("selector1, selector2, #test")). Hard to tell, based on how he's written it. — talemyn
– talemyn, Commented Mar 6, 2015 at 22:14
@SterlingArcher The first argument will be a selector and the second one the context. — Ram
– Ram, Commented Mar 6, 2015 at 22:14
The specifics of the selector don't really matter, but I saw it as multiple selectors. That's why I didn't post as answer lol — Sterling Archer
– Sterling Archer, Commented Mar 6, 2015 at 22:15

RickyTomatoes · Accepted Answer · 2015-03-07 05:01:45Z

1

EDIT:

Yes, but not exactly like that. If each of your $selector variables represents a collection of a single DOM element, you can rewrite your code like this:

$([$selector1[0], $selector2[0], document.getElementByID('test')])

And you will be able to add a class to all of those elements at once without .each()

edited Mar 7, 2015 at 5:01

answered Mar 6, 2015 at 22:18

RickyTomatoes

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3 Comments

goodpixels Over a year ago

Yeah, but I still need to use $.each - I wanted to specifically avoid this.

Ifedi Okonkwo Over a year ago

I guess you've got a little typo. The opening square bracket is missing after $(. Great answer, though!

RickyTomatoes Over a year ago

@IfediOkonkwo Thanks for pointing that out, missed it in my haste to type

Jonathan Lonowski · Accepted Answer · 2015-03-06 22:18:39Z

1

I don't believe it supports it within a single function call.

But, you could chain calls to .add():

$().add($selector1).add($selector2).add(document.getElementById('test'))

When given an Array, jQuery() generally expects the contents to already be Elements. Otherwise, it treats it as using the jQuery(object) overload, leaving the strings as is.

console.log( $(['#foo', '#bar']).get() ); // [ '#foo', '#bar' ]

answered Mar 6, 2015 at 22:18

Jonathan Lonowski

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1 Comment

goodpixels Over a year ago

While this is not really what I'm looking for, it's pretty close. Thanks.

Ifedi Okonkwo · Accepted Answer · 2015-03-06 22:21:20Z

0

I think the short answer is "no". JQuery does not expect array as selector.

Its either a comma-separated string list, or a single DOM object or a jQuery object.

A second argument is treated as the "parent" container within which to conduct the search.

If you insist on collecting several objects into one, an option is to do this:

$($selector1).add( $selector2).add(document.getElemetByID( 'test' ))

edited Mar 6, 2015 at 22:21

answered Mar 6, 2015 at 22:14

Ifedi Okonkwo

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