When the user clicks on an input button where "cancel" is added to the
class attribute, the form submits as expected, but when the the input
button is submitted by script like this, $("input
[name='Btn_Cancel']").submit(), then the validation fires as if the
class="cancel" is not there.
Any idea w
Exactly. $("div.image")[4] gives the raw dom element from the jQuery
collection, and not a jQuery "wrapped" element like the OP was expecting
(thus the .show() function isn't available).
Re-wrapping the element like so: $($("div.image")[4]).show() would work. Of
course you'd be better off using
You mentioned the validate plugin: last I looked it was still using xpath
style '@' for attribute selectors
( e.g. $(inp...@name=some]), which has been removed from 1.3 in favour of
plain [name=some]. Could be your problem?
jQuery.Nikola wrote:
>
>
> Well, that was really strange. The prob
I can't help but think that supporting Firefox and Safari is a but of a
waste. Surely a developer would be better off doing something like this:
-moz-border-radius:5px;
-webkit-border-radius:5px;
border-radius:5px;
...and using curvy corners in a similar manner to png or
If you aren't sure perhaps you can "return true" in the $.get callback
somehow?
wswilson wrote:
>
>
> My code:
>
> ---in the script---
> $('a').click(function() {
> $.get('/long_running_page');
> return true;
> }
>
> ---on the page---
> http://www.otherserver.com click here
>
>
Does setting the "contentType" work?
http://docs.jquery.com/Ajax/jQuery.ajax#options
Chris Bailey-4 wrote:
>
>
> I'm using jQuery 1.2.6, and I can't seem to get jQuery to set the X-
> Requested-With when it makes AJAX calls. I saw somewhere that this
> may have to do with cross-site issues i
Perhaps you want to use .queue() to see if there is an animation already in
progress before triggering a new one.
untested, but something like this in the hover functions might be the go:
if ($(this).queue("fx").length < 1) ...etc
http://docs.jquery.com/Effects/queue for more info
next wro
e().end();
That will remove that span from the document, and leave you only with its
containing div...
duck! wrote:
>
> $("div:not(span)"); will return all the divs in the document that are not
> spans... all of them obviously
>
> You want all the divs that don't (:not
$("div:not(span)"); will return all the divs in the document that are not
spans... all of them obviously
You want all the divs that don't (:not()) contain (:has()) spans.
$("div:not(:has(span))");
hope this helps.
jquertil wrote:
>
>
> according to jquery documentation I should be able to s
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