[jQuery] Re: Running a loop for array numbers

[Jonathan Sharp]

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.
>