how about
$("div:has(a)").each(function() {
var $div = $(this); //Here's the div
var $a = $div.find("a"); //Here's the link
})
On Apr 3, 9:56 am, patrickk <patr...@vonautomatisch.at> wrote:
> when using something like:
>
> $('div a').each(function() {
> // what´s the easiest way to get the "DIV" here?
>
> }
>
> "div" doesn´t need to be a parent of "a". so, I need to know the exact
> location (DOM-wise) of my links to get the "div". I can somehow
> remember that there´s an extremely easy way to get the "div", but
> unfortunately I can´t remember ...
>
> thanks,
> patrick
