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#8937 closed bug (invalid)

Opened April 21, 2011 04:41PM UTC

Closed April 22, 2011 01:41PM UTC

wrapAll() Doesn't Expose Wrapper

Reported by: matthew@exanimo.com Owned by: matthew@exanimo.com
Priority: low Milestone: 1.next
Component: manipulation Version: 1.5.2
Keywords: Cc:
Blocked by: Blocking:
Description

After using wrapAll() to create a wrapper, there's no way to get a reference to that wrapper (without reselecting, e.g. with an id).

var wrapper = $('<div />');
$('.some-consecutive-elements').wrapAll(wrapper);
wrapper.append('<span>Some new text</span>'); // Will not work because the "wrapper" variable is not actually a reference to the wrapping element. You must perform a new selection to get the wrapping element.
Attachments (0)
Change History (2)

Changed April 21, 2011 04:44PM UTC by timmywil comment:1

component: unfiledmanipulation
owner: → matthew@exanimo.com
priority: undecidedlow
status: newpending
type: enhancementbug

Thanks for taking the time to contribute to the jQuery project! Please provide a reduced test case on http://jsFiddle.net that reproduces the issue experienced to help us assess your ticket.

Additionally, test against the jQuery (edge) version to ensure the issue still exists.

Changed April 22, 2011 01:41PM UTC by dmethvin comment:2

resolution: → invalid
status: pendingclosed

That's the way the API is documented to work. Just like .clone() selects the clones, not the original elements.