On Oct 2, 6:09 pm, Wizzud <[EMAIL PROTECTED]> wrote:
> No, you can't check for parent conditions with is(). You can test descendants
> to a certain extent, but not parents.
> The same goes for filter() - unless you use filter(function), in which case
> you do more or less what you like!
> Also, using hierarchical selectors as filters won't work - they're
> selectors, not filters.
Ah - is the distinction made somewhere that I missed? Since "filters"
are just a subset of the "selectors" it wasn't clear to me which types
of expressions would work.
> The third of your filter() tests ...
> alert ( $('div').filter('div:not(.test) div').size() ); // Alerts 3
> ... is treating the filter as 2 separate filters and adding the results, eg
> filter('div:not(.test)') + filter('div').
Which is fine, I guess that functionality is acceptable as long as I
understand that "filters" are different than "selectors". I just won't
use that syntax and expect the results I wanted. :)
> Is there a better way than...?
> if (o.parents().filter('div.unwantedclass').size()==0) { ... }
> It really depends on what you're using the 'o' collection for. The obvious
> alternative is not to put into 'o' anything that has a parent of class
> 'unwanted' to start with, eg o = $('div:not(.unwantedclass)>*');
Unfortunately I'm doing this check on an event handler and checking on
what kind of element is clicked to determine what to do. Due to the
size of the page, using selectors on page load to attach the correct
event actions caused way too much slow down (over 10 seconds, page is
> 400k from an internal webapp). So I'm catching the event at the
container and then determining which element was clicked so I can take
appropriate action.
> If you can't do that, then there's nothing wrong with what you have, but a
> filter alternative (>=v1.2) could be...
> o.filter(function(){return !$(this).parent().is('unwantedclass');});
True, but I'm needing to checking for the class on any element in the
parent chain, not just the immediate parent. So I guess the way I'm
doing it will suffice. Thanks for your help!
Matt Kruse
