Using two variables as selectors to call a function (jQuery)

Question

Is it possible, after declaring two vars, to use them together to call a function in jQuery?

like:

var test1 = $('#mytest1');
var test2 = $('#mytest2');

test1,test1.fadeOut(100);

I would recommend you to use JSLint to help you to find common mistakes in scripts, If you try to do something like var1, var2.method(), it will say Expected an assignment or function call and instead saw an expression — abestrad
– abestrad, Commented Mar 7, 2019 at 18:25

Blue · Accepted Answer · 2019-03-07 18:31:01Z

4

Yes. You can just include the selectors within quotes to the $ function:

$('#mytest1, #mytest2').fadeOut(2000);

#mytest1, #mytest2 {
  width: 50px;
  height: 50px;
  margin: 5px;
  background-color: blue;
}

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<div id="mytest1"></div>
<div id="mytest2"></div>

If you wish to use the existing variables you've already declared, you can use .add() as noted in this answer:

var test1 = $('#mytest1');
var test2 = $('#mytest2');

$(test1).add(test2).fadeOut(2000);

#mytest1, #mytest2 {
  width: 50px;
  height: 50px;
  margin: 5px;
  background-color: blue;
}

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<div id="mytest1"></div>
<div id="mytest2"></div>

answered Mar 7, 2019 at 18:19

Blue

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1 Comment

You should "normally hide" the first snippet and show the second...;) +1

Taplar · Accepted Answer · 2019-03-07 18:24:43Z

2

$('#mytest1, #mytest2').fadeOut();

If you already have both in two separate variables, you can combine them. Ref. http://api.jquery.com/add/

test1.add(test2).fadeOut();

answered Mar 7, 2019 at 18:24

Taplar

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Whoops... You actually replied the same as me faster! I did not see. ;)

That's perhaps what I was looking for :) Thanks