jQuery inArray() locating HTML element

Question

Say,I have 5 paragraphs in the page.

if I execute:

p_array=$('p');
second_p=$('p:eq(1)');
$.inArray(second_p,p_array);

I get -1. Any explanation?

second_p is a jQuery object, and p_array is an array-like object which contains DOM nodes... Therefore, p_array obviously does not contain second_p. — Šime Vidas
– Šime Vidas, Commented Jun 13, 2011 at 0:36
Why not second_p = p_array.eq(1);? You're doing two look-ups unnecessarily. — Šime Vidas
– Šime Vidas, Commented Jun 13, 2011 at 0:38

Lightness Races in Orbit · Accepted Answer · 2011-06-13 00:32:35Z

4

Neither p_array nor second_p are arrays.

They are jQuery objects.

More specifically, p_array is a jQuery object containing a set of 5 DOM nodes. second_p is a jQuery object containing a set of 1 DOM node.

$.inArray can function on these jQuery sets of nodes, but you can't compare a set against a set.

If you extract that one DOM node using the array subscript operator (jQueryObj[i]), then you're no longer comparing a set against a set:

var p_array=$('p');
var second_p=$('p:eq(1)');
alert($.inArray(second_p[0], p_array)); // result: 1

See a live demo here.

answered Jun 13, 2011 at 0:32

387k77 gold badges670 silver badges1.1k bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

@xiaohan2012: Yea, we like it :)

Sophie Alpert · Accepted Answer · 2011-06-13 00:33:03Z

1

Both p_array and second_p are array-like jQuery objects. You want to extract the actual DOM node for the <p> and test it like:

$.inArray(second_p[0], p_array);

That call returns 1 for me, as expected.

answered Jun 13, 2011 at 0:33

Sophie Alpert

144k37 gold badges229 silver badges244 bronze badges