Comments on: .parents()

By: Snail

Snail — Wed, 29 Dec 2010 16:15:02 +0000

When you use parents() on an element in iframe, the iframe is not in parents list.

By: Snail

Snail — Wed, 29 Dec 2010 08:15:02 +0000

When you use parents() on an element in iframe, the iframe is not in parents list.

By: d.u.a.l

d.u.a.l — Sat, 18 Dec 2010 06:31:03 +0000

like somebody said before, closest is your choice

By: Adam Crabtree

Adam Crabtree — Mon, 29 Nov 2010 12:24:13 +0000

.closest(), http://api.jquery.com/closest/, is what you want.

By: Antonio Trapani

Antonio Trapani — Sat, 27 Nov 2010 07:10:59 +0000

Just read .closest() documentation, it comes in 1.3. Thanks

By: Karl Swedberg

Karl Swedberg — Sat, 27 Nov 2010 01:41:49 +0000

No, unfortunately, .parents() will still traverse to the root element. You could use .closest(), but keep in mind that it starts with the element(s) in the jQuery set rather than with the parent element(s).

By: Antonio Trapani

Antonio Trapani — Mon, 22 Nov 2010 06:33:56 +0000

I was wondering how I can stop the parents(selector) method after the first occurrence of the selector is found… then I realized that using the pseudo-selector :first could help. Am I right? Does parents(selector:first) stops after the first element is found, without traversing the DOM up to the root element?

By: Karl Swedberg

Karl Swedberg — Fri, 01 Oct 2010 23:51:33 +0000

It traverses up to the root element and filters the result.

By: Gene

Gene — Thu, 30 Sep 2010 07:00:00 +0000

What about using $(this).parents(‘.second-level-parent:eq(0)’); ? Is jQuery smart enough to stop after the first one, or does it still traverse the entire tree and then filter the results?

By: Karl Swedberg

Karl Swedberg — Sun, 29 Aug 2010 15:04:42 +0000

I assume you meant to compare $(this).parent().parent() to $(this).parents('.second-level-parent'). Note the “s” in the second. It depends on how deeply nested your elements are. .parents() will traverse all the way up the DOM to the root element (typically ) and check for a match (if you've provided a filter) on each element. So it's potentially, but not necessarily, much slower.