adding newly created dom elements to an empty jQuery object

Question

Why doesn't this work in jQuery 1.4.2?


var $list = $([]);
for(var i=0; i<50; i++) {
    $list.add( $('<div/>', { id: 'jake', class: 'test' }).data('test', { hi: 'hello' }) );
}
alert($list.size()); // 0

Thanks!

Correct explanation given at : [jQuery add elements to empty selection?][1] [1]: stackoverflow.com/questions/7533929/… — mohanrajt
– mohanrajt, Commented Apr 18, 2013 at 7:00

Rahen Rangan · Accepted Answer · 2014-04-06 06:34:45Z

11

Pointing back the reference list again works for me; e.g. $list = $list.add( $('<div/>') );

var $list = $([]);
for(var i=0; i<50; i++) {
    $list=$list.add( $('<div/>', { 'id': 'jake'+i, 'class': 'test' }).data('test', { hi: 'hello' }) );
}
alert($list.size()); // 50

edited Apr 6, 2014 at 6:34

answered May 19, 2012 at 7:11

Rahen Rangan

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1 Comment

Michael come lately Over a year ago

Upvote for solving my problem, but I wish there was a better way (read: .append()) to add items to an empty set. Shog9's answer to a similar question explains why there isn't.

x1a4 · Accepted Answer · 2010-05-14 05:35:16Z

4

Why add doesn't work I don't know, but you can replace it with push due to jQuery being an Array-like object, which should do what you want.

answered May 14, 2010 at 5:35

x1a4

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5 Comments

taber Over a year ago

awesome, i thought i tried that too, but apparently not! thanks.

taber Over a year ago

it looks like in order to use .clone(true) on $list, my $list array needs to be a dom node (eg: $('<div />') but i don't want a containing div, i just want a list of dom nodes with no parent. (like an array!) is that possible?

taber Over a year ago

it's not the end of the world if i have to have a containing div i guess. it's just extra junk in the dom. oh well.

x1a4 Over a year ago

In your case, because each div is created separately before being pushed onto $list, each element in $list is actually its own separate jQuery object, so THOSE are what you need to call clone on: clones = $list.map(function () { return this.clone(true); }); and you should be good to go.

x1a4 Over a year ago

and no, you definitely won't need the containing div :)

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