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#4316 closed bug (worksforme)

Opened March 09, 2009 07:14PM UTC

Closed October 28, 2009 02:04PM UTC

Safari 4 Fade Issues

Reported by: whoughton Owned by:
Priority: minor Milestone: 1.4
Component: effects Version: 1.3.2
Keywords: Cc:
Blocked by: Blocking:
Description

http://www.mixicon.com/tools/jqtest.html

Using fadeIn, FadeOut, fadeTo all create an issue in the Safari 4 beta, I'm not sure if this is Safari's problem or jQuery's.

On that page, Safari shows an error of:

TypeError: Result of expression '(o.fx.step[this.prop]||o.fx.step._default)' [0] is not a function.

Attachments (0)
Change History (3)

Changed March 09, 2009 08:33PM UTC by fALk comment:1

I can second this bug. Seems to work with build 5528.16 but the latest nightly webkit build 5528.16, r41521 shows above behavior.

Changed March 10, 2009 03:18AM UTC by whoughton comment:2

Submitted this to the webkit bugs as well, considering that it may be a squirrelfish bug (or whatever it's called now).

https://bugs.webkit.org/show_bug.cgi?id=24471

Changed October 28, 2009 02:04PM UTC by davidserduke comment:3

description: \ http://www.mixicon.com/tools/jqtest.html \ \ Using fadeIn, FadeOut, fadeTo all create an issue in the Safari 4 beta, I'm not sure if this is Safari's problem or jQuery's. \ \ On that page, Safari shows an error of: \ TypeError: Result of expression '(o.fx.step[this.prop]||o.fx.step._default)' [0] is not a function.http://www.mixicon.com/tools/jqtest.html \ \ Using fadeIn, FadeOut, fadeTo all create an issue in the Safari 4 beta, I'm not sure if this is Safari's problem or jQuery's. \ \ On that page, Safari shows an error of: \ TypeError: Result of expression '(o.fx.step[this.prop]||o.fx.step._default)' [0] is not a function.
resolution: → worksforme
status: newclosed

Seems to work in the release version of Safari 4 and the webkit ticket indicates they made a fix so closing on that basis.