How can I wrap two jquery objects into one?

Question

I have two functions.

1

function foo(elems){ elems.each(function(){ }); } foo($('selector1,selector2')); // calling foo

2

function bar(elem1,elem2){ // some logic } bar($('selector1'),$('selector2)); // calling bar

However, now I need to use foo() inside bar().

function bar(elem1,elem2){ foo(?); // I'm stuck here. I want to pass both elem1 and elem2 as single jquery object }

So I tried $(elem1,elem2) but it only includes elem1. So what should I do, so that I can pass elem1 and elem2 as a single jquery object which I'll iterate over in foo() by .each()

CodingIntrigue · Accepted Answer · 2014-04-12 15:08:02Z

2

You can use .add:

foo(elem1.add(elem2));

To combine the results of both selectors

answered Apr 12, 2014 at 15:08

CodingIntrigue

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3 Comments

Amit Joki Over a year ago

+1 how could I forget .add()? Thanks will accept it as soon as I can :)

CodingIntrigue Over a year ago

It's an easy one to forget when you're writing selectors all the time :-)

Amit Joki Over a year ago

You see the irony here? My question is being upvoted, but I've reached 200 limit already lol

attila · Accepted Answer · 2014-04-12 15:08:13Z

You can use the jQuery "add" method, like this

$(elem1).add(elem2); //this now contains both elements

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