Hi Johnee,
Here's one approach (untested):
var $e = $('.equipment');
$.each(['1', '2', '5', 'x'], function(i, k) {
$e .find('a.i-right' + k)
.filter(function(i) {
return ( i > 0 && i < 9 );
})
.hide()
.end();
});
Cheers,
-Jonathan
On Wed, Jul 9, 2008 at 3:55 AM, JohneeM <[EMAIL PROTECTED]> wrote:
>
> Thanks guys much appreciated.
>
> I tried the last one :
>
> $('.equipment a.i-right1').filter(function(i) {
> return ( i > 0 && i < 9 );
>
> }).hide();
>
> And it worked.
>
> For say
>
> $('.equipment a.i-right1')
> $('.equipment a.i-right2')
> $('.equipment a.i-right5')
> $('.equipment a.i-rightx')
>
> How would i put that into a loop?
>
> Thanks.
>
> On Jul 8, 10:16 pm, "Jonathan Sharp" <[EMAIL PROTECTED]> wrote:
> > Hi Johnee,
> >
> > Another approach would be:
> >
> > $('.equipment a.i-right1').each(function(i) {
> > if ( i > 0 && i < 9 ) {
> > $(this).hide();
> > }
> >
> > });
> >
> > or this would work too:
> >
> > $('.equipment a.i-right1').filter(function(i) {
> > return ( i > 0 && i < 9 );
> >
> > }).hide();
> >
> > Cheers,
> > -Jonathan
> >
> > On Tue, Jul 8, 2008 at 12:00 PM, JohneeM <[EMAIL PROTECTED]> wrote:
> >
> > > Hi guys how can i run this in a single statement without manually
> > > putting in the numbers?
> >
> > > $('.equipment a.i-right1:eq(1)').hide();
> > > $('.equipment a.i-right1:eq(2)').hide();
> > > $('.equipment a.i-right1:eq(3)').hide();
> > > $('.equipment a.i-right1:eq(4)').hide();
> > > $('.equipment a.i-right1:eq(5)').hide();
> > > $('.equipment a.i-right1:eq(6)').hide();
> > > $('.equipment a.i-right1:eq(7)').hide();
> > > $('.equipment a.i-right1:eq(8)').hide();
> >
> > > Note that im skipping the first array [0] due to wanting it to show.
> >
> > > Also this one :
> >
> > > $('.performance-parts #content-container
> > > a.lightbox:eq(8)').lightBox();
> > > $('.performance-parts #content-container
> > > a.lightbox:eq(9)').lightBox();
> >
> > > They need to be in their own array as i want to run a seperate
> > > lightbox for each image, is this possible?
> >
> > > Thanks.
>
