ansaurus

Question

Answer 1

A:

$(a).find("input")

Josh Stodola 2009-08-12 14:41:35

If `a` comes from previous jQuery selection there is no need to wrap it around `$()`.

RaYell 2009-08-12 14:45:44

find will not work as it searches descendant elements, these are all level 1 elements.

maxp 2009-08-12 14:47:19

I started answer with filter function, then edited and changed to find, because your question was not clear.

Josh Stodola 2009-08-12 15:08:28

Answer 2

+1 A:

$('input', a);

RaYell 2009-08-12 14:42:44

This doesnt do anything.

maxp 2009-08-12 14:48:28

This searches input tags among the descendants of `a` and it works.

RaYell 2009-08-12 14:56:08

I think it's $('input', 'a');

rball 2009-08-12 15:20:40

No, it's not. `a` is a jQuery object as stated in the question.

RaYell 2009-08-12 15:29:53

Answer 3

+4 A:

a.filter('input').each(function() {
    alert('My name is ' + $(this).attr('name'));
});

To just get a selection from the current selection this way:

var $inputs = a.filter('input');

You can even comma separate selectors:

var $els = a.filter('input, .fooMonger, #something');

karim79 2009-08-12 14:44:30

Thanks! I woulod give you more upvotes if i could.

maxp 2009-08-12 15:05:44

Jquery: Selection within a selection.