Hi,
I relative new to JQuery but I have some code that works fine in
Firefox, IE and Opera, but seem to be having a problem with Chrome and
Safari.
Here is part of the code:
$(form#quiz_form_1).submit(function(){
alert(This is testing);
$.post(question_result.asp,{
You should remove the @ from your selectors, they're invalid.
--John
On Thu, Apr 2, 2009 at 12:10 PM, ale alejandr...@gmail.com wrote:
Hi,
I relative new to JQuery but I have some code that works fine in
Firefox, IE and Opera, but seem to be having a problem with Chrome and
Safari.
Sure!
Created http://dev.jquery.com/ticket/4058
Hope there is some easy patch, if not, I will regret to 1.2.6 inmediately :(
2009/2/3 John Resig jere...@gmail.com
That's odd. Could you file a bug on this?
http://dev.jquery.com/newticket
Thanks!
--John
On Tue, Feb 3, 2009 at 10:39
Try this:
var containerSelecteds = function() {
return $('ul.selected', container);
};
I've had pretty good luck using that syntax instead of using .find()
or .children(). It might solve the problem on safari (as I think the
reason I stopped using find() or children() was that the above
Thanks for the fast response.
Unfortunately it doesn't work :(
2009/2/3 Eric Garside gars...@gmail.com
Try this:
var containerSelecteds = function() {
return $('ul.selected', container);
};
I've had pretty good luck using that syntax instead of using .find()
or .children(). It
That's odd. Could you file a bug on this?
http://dev.jquery.com/newticket
Thanks!
--John
On Tue, Feb 3, 2009 at 10:39 AM, Javier Martinez ecentin...@gmail.com wrote:
I'm creating a component for an application I'm developing and I have
upgraded jquery to the last version to get it's speed
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