whre i have to put this breakpoint in firebug?
On Apr 8, 12:45 pm, donb wrote:
> I would breakpoint 'deeper', down in the ajax code such as where
> ajaxStart is called. Then check the stack to see how you got there.
>
> On Apr 8, 4:53 am, Skyblaze wrote:
>
> >
0700
> > Subject: [jQuery] Re: Jquery makes two ajax requests in one!!
> > From: marcomenozz...@gmail.com
> > To: jquery-en@googlegroups.com
>
> > this is very very strange...any ideas?
>
> > On Apr 8, 1:52áam, Skyblaze wrote:
> > > ok i did that and i
this is very very strange...any ideas?
On Apr 8, 1:52 am, Skyblaze wrote:
> ok i did that and i have only one alert box so the event is fired
> once. So what is that fires two same ajax get request?
>
> On Apr 8, 1:00 am, James wrote:
>
> > Try adding an alert or conso
ent is
> actually being called once or twice.
>
> On Apr 7, 12:23 pm, Skyblaze wrote:
>
> > I have no other id with that name and also with firebug suspended it
> > is the same.always the same ajax/get request...two identical ajax
> > get request one after the other
>
> I had some code left over from experimentation on a page that
> I failed to clear off and every time a certain ajax function ran,
> it did so twice!
>
> Rick
>
> -Original Message-
> From: jquery-en@googlegroups.com [mailto:jquery...@googlegroups.com] On
>
> Be
i have the same strange problem
On Apr 7, 10:38 am, EnvyGeeks wrote:
> It seems that either I'm doing something horribly wrong or something
> because everytime I do anAjaxcall using $.load() it causes adouble
> query on the page.
>
> example:
> $("div#example a").click(function(){
> $.ajax_u
I have a strange problem. I have to do an ajax request after a select
box changes (change event) so i have the following code:
$('#comprensori_id').change(function() {
var comprensorio_id = $(this).val();
$.ajax({
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