what does wrapping a jQuery object with another `$` mean?

Asked Jan 31 '16 at 19:51

Active Jan 31 '16 at 19:51

Viewed 31 times

I have this code:

$('select[name="dropdown"]').change(function() {

var item = this.selectedIndex;

why does selectedIndex return undefined when I wrap $(this)

I thought if this is already an jQuery object, than nothing will happen when I wrap with redundant $

Is there need to check something before wrapping with $ ?

what does exactly it means to wrap with $ ?

asked Jan 31 '16 at 19:51

Elad Benda

4

`this` inside the handler is the **native DOM node** not a jQuery object. Native DOM nodes have a `selectedIndex` property, jQuery objects do **not**! – adeneo Jan 31 '16 at 19:53
4

`this` == DOM element `$(this)` == jQuery object, Also - `$(this)[0] === this` – Alon Eitan Jan 31 '16 at 19:53
`this` is a javascript thing, `$(this)` was specifically made for jQuery. – Jan 31 '16 at 19:57
What will happen if I do $($(this)) meaning I wrap jquery object with jQuery again? – Elad Benda Feb 01 '16 at 08:50
As you suspected, nothing will happen. – Clayton Leis Feb 01 '16 at 10:18

0 Answers0