#3895 closed enhancement (fixed)
Using offsetWidth for jQuery selectors (especifically the :visible selector)
Reported by: | matheusalmeida | Owned by: | john |
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Priority: | minor | Milestone: | 1.3.2 |
Component: | selector | Version: | 1.3.1 |
Keywords: | visible | Cc: | |
Blocked by: | Blocking: |
Description
As discussed at the jQuery Development group: http://groups.google.com/group/jquery-dev/browse_thread/thread/e18b399b36d03f4f
A simpler and more officient way to determine if a element is visible at the document is to test its offsetWidth or offsetHeight parameter values.
It discards the parents verification and treat as hidden not just the elements with display:none or visibility:hidden but also those that has zero-valued width or height.
Change History (2)
comment:1 Changed 14 years ago by
comment:2 Changed 14 years ago by
Milestone: | 1.3.1 → 1.3.2 |
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Resolution: | → fixed |
Status: | new → closed |
Version: | 1.2.6 → 1.3.1 |
Fixed in SVN rev [6210].
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I've had a similar problem with version 1.3.1 !
It was found in UI/Effects/Transfer. When target element with display:none ,the function offset() return zero-valued top and left.
I've had found a temporary way to solve the problem.
/ jQuery.iUtil = {
..... };
These codes from http://interface.eyecon.ro ,It might be helpful to settle this matter finally.
Complete
/
jQuery.iUtil = {
};