[jQuery] Re: possible bug in each()? moving from 1.2.3 to 1.2.6

2008-08-26 Thread mausch
Thanks Ariel! On Aug 26, 3:43 pm, Ariel Flesler <[EMAIL PROTECTED]> wrote: > You can either update Prototype/Scriptaculous to 1.8.x or try the > latest jQuery revision on the trunk. That should be 'patched'. > > -- > Ariel Fleslerhttp://flesler.blogspot.com > > On Aug 25, 1:59 pm, mausch <[EMAIL

[jQuery] Re: possible bug in each()? moving from 1.2.3 to 1.2.6

2008-08-26 Thread Ariel Flesler
You can either update Prototype/Scriptaculous to 1.8.x or try the latest jQuery revision on the trunk. That should be 'patched'. -- Ariel Flesler http://flesler.blogspot.com On Aug 25, 1:59 pm, mausch <[EMAIL PROTECTED]> wrote: > Nevermind, it was executing prototype's each() (I was also includi

[jQuery] Re: possible bug in each()? moving from 1.2.3 to 1.2.6

2008-08-25 Thread mausch
Nevermind, it was executing prototype's each() (I was also including prototype on the same page) instead of jQuery's. Why this changed from 1.2.4 to 1.2.6 (everything else being equal) remains a mystery to me. On Aug 25, 1:12 pm, mausch <[EMAIL PROTECTED]> wrote: > I'm having the same problem her

[jQuery] Re: possible bug in each()? moving from 1.2.3 to 1.2.6

2008-08-25 Thread mausch
I'm having the same problem here. If I execute the following in a firebug console in one of our pages: jQuery('input').each(function() {console.log(this)}) With 1.2.4 I get the individual elements: ... With 1.2.6 I get one array containing all matching elements: [input, input#InputSearch,

[jQuery] Re: possible bug in each()? moving from 1.2.3 to 1.2.6

2008-07-02 Thread Brian J. Fink
Is that the correct way to spell it? Maybe all lowercase would work. (I don't know. I use $ myself.) On Jul 2, 6:48 pm, darren <[EMAIL PROTECTED]> wrote: > Hi > > Im trying to use 1.2.6, moving from 1.2.3.  I have a working function > in 1.2.3 but it is not working with the updated jquery library

[jQuery] Re: possible bug in each()? moving from 1.2.3 to 1.2.6

2008-07-02 Thread Michael Geary
Try stepping *into* the .each() function instead of *over* it. That way you can see how it works and how it calls down into your callback function. There's not much to go wrong in .each() itself, so the error you're getting is probably coming from your callback function. You can set "Break on all