[jQuery] Re: jQuery works in one place but not another
When you call $(document).ready, the $ is equal to jQuery, so it works.
You're passing it a callback function to be executed when the DOM is ready.
By that time another library has loaded and overwritten the $. For possible
work-arounds see
http://docs.jquery.com/Using_jQuery_with_Other_Libraries
Especially the last few bullet points.
- Richard
On Fri, Jul 11, 2008 at 2:19 PM, Josh <[EMAIL PROTECTED]> wrote:
>
> I just tried this change:
>
>
>
> $(document).ready(function () {
>alert('this works');
> $('h2').fadeOut('slow');
> });
>
>
> Same problem: "$ is not a function." The alert works, but the
> fadeOut() doesn't. Why would it work for $(document) but not the
> other?
>
> On Jul 11, 11:38 am, Josh <[EMAIL PROTECTED]> wrote:
> > I'm observing some strange behavior with jQuery 1.2.6 today. I've got
> > this in my
> >
> >
> >
> > $(document).ready(function () {
> > alert('this works');
> > });
> >
> > function showFilterCalendar() {
> > $('h2').fadeOut('slow');
> >
> > // don't follow the link
> > return false;
> > }
> >
> >
> > When the page loads, I get the alert box saying "this works."
> > However, when I click a link like > onclick="javascript:return showFilterCalendar()">test, I get the
> > infamous "$ is not a function" error.
> >
> > Ideas?
>
[jQuery] Re: jQuery works in one place but not another
Ok, I think I found a way around the problem. Apparently one of the
other scripts my institution uses was overwriting $ or something...
Including the jQuery library at the end of the HTML file (while not
proper) seems to get around it.
On Jul 11, 11:38 am, Josh <[EMAIL PROTECTED]> wrote:
> I'm observing some strange behavior with jQuery 1.2.6 today. I've got
> this in my
>
>
>
> $(document).ready(function () {
> alert('this works');
> });
>
> function showFilterCalendar() {
> $('h2').fadeOut('slow');
>
> // don't follow the link
> return false;
> }
>
>
> When the page loads, I get the alert box saying "this works."
> However, when I click a link like onclick="javascript:return showFilterCalendar()">test, I get the
> infamous "$ is not a function" error.
>
> Ideas?
[jQuery] Re: jQuery works in one place but not another
I just tried this change:
$(document).ready(function () {
alert('this works');
$('h2').fadeOut('slow');
});
Same problem: "$ is not a function." The alert works, but the
fadeOut() doesn't. Why would it work for $(document) but not the
other?
On Jul 11, 11:38 am, Josh <[EMAIL PROTECTED]> wrote:
> I'm observing some strange behavior with jQuery 1.2.6 today. I've got
> this in my
>
>
>
> $(document).ready(function () {
> alert('this works');
> });
>
> function showFilterCalendar() {
> $('h2').fadeOut('slow');
>
> // don't follow the link
> return false;
> }
>
>
> When the page loads, I get the alert box saying "this works."
> However, when I click a link like onclick="javascript:return showFilterCalendar()">test, I get the
> infamous "$ is not a function" error.
>
> Ideas?
