An array of elements in jQuery

Question

Markup code:

<div id="elements">
    <div>
        <a href="#">text</a>
        <a href="#">text</a>
        <a href="#">text</a>
    </div>
    <div>
        <a href="#">text</a>
        <a href="#">text</a>
        <a href="#">text</a>
    </div>
    <div>
        <a href="#">text</a>
        <a href="#">text</a>
    </div>
</div>

Please tell me how can I get an array of all elements of the div, so that later, it is possible to address an array of options?

Such as:

divs[0]  
links[1]

Take a look here: stackoverflow.com/q/247023/844726 Quite a few options. — swatkins
– swatkins, Commented Oct 1, 2011 at 20:22

Paul Schwarz · Accepted Answer · 2017-02-07 15:25:35Z

16

See https://api.jquery.com/toArray/

$( "li" ).toArray()

This works great in cases like

$( "li" ).toArray().each( function(li) {
  console.log('do something with this list item', li);
})

or if you really want

$( "li" ).toArray()[0]

answered Feb 7, 2017 at 15:25

Paul Schwarz

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1 Comment

png Over a year ago

This is the correct answer. The accepted answer is outdated.

Shef · Accepted Answer · 2011-10-01 20:28:17Z

9

Demo

var divs = $('#elements div');
var links = $('#elements div a');

If you want the DOM elements, then you can access them with the array style like divs[0] or links[2].
If you want to get the specific jQuery object, you can access them with divs.eq(0) or links.eq(1).

edited Oct 1, 2011 at 20:28

answered Oct 1, 2011 at 20:22

Shef

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3 Comments

SirRoland Over a year ago

I get an array of links for each div separately? $ ('# elements div [0] a'); like - something like that, just to correct it. ) I need to do the sorting. :)

Shef Over a year ago

@SirRoland Yes, you can do $('#elements div:eq(0) a');.

SirRoland Over a year ago

if I write: var links=$('#elements div:eq(0) a').length; alert(links); // = 0

meder omuraliev · Accepted Answer · 2011-10-01 20:30:47Z

8

$('#elements div').eq(0) // first div

$('#elements div a').eq(1) // second anchor in group of divs

Without the .eq() you have a collection of elements, the .eq just singles the elements out.

edited Oct 1, 2011 at 20:30

answered Oct 1, 2011 at 20:21

meder omuraliev

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Comments

mowwwalker · Accepted Answer · 2011-10-01 20:23:56Z

wrapper = document.getElementById('elements');
divs = wrapper.getElementsByTagName('div');
links = [];
for (i = 0; i < divs.length; i++) {
    links[i] = divs[i].getElementsByTagName('a');
}

divs will now be an array of the divs inside of your "elements" div. Links is now a two dimensional array, where the first index refers to the div inside your "elements" div, and the second index refers to the "a" tags inside that div.

links[2][1]

will refer to the element denoted below:

<div id="elements">
    <div>
        <a href="#">text</a>
        <a href="#">text</a>
        <a href="#">text</a>
    </div>
    <div>
        <a href="#">text</a>
        <a href="#">text</a>
        <a href="#">text</a>
    </div>
    <div>
        <a href="#">text</a>
        <a href="#">text</a>  //this one
    </div>
</div>

maxijb · Accepted Answer · 2011-10-01 20:38:30Z

1

Or also, and I believe this is more accurately what you asked for:

$("#elements div:eq(0) a:eq(1)")

answered Oct 1, 2011 at 20:38

maxijb

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Comments

Nikola K. · Accepted Answer · 2012-07-31 14:34:38Z

0

$("#elements a")

$("div a")

or $("a")

jQuery Selectors

edited Jul 31, 2012 at 14:34

Nikola K.

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answered Oct 1, 2011 at 20:21

Mirko

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