You can loop over the colors also just by using the modulo operator
instead:
$('a').each(function(i){
i = ++i % 4;
var color =
i == 1
? 'red' : //first 4
i == 2
? 'blue' : //second group of 4
i == 3
? 'yellow' : // third group
'lime'; //fourth group and so on
$(this).css('backgroundColor', color);
});
On 23 jan, 15:38, devilmike <devilm...@gmail.com> wrote:
> Awesome Ricardo, thanks! I guess the only issue I have is that I'll
> never know how many "sets of4" I'll be dealing with, and i apologize
> for not explaining myself very well in my example. Basically for each
> "set", I want to run the same function. This is what I came up with.
> It works, but I'm a bit concerned about the amount of looping going
> on. Your code seems much cleaner...
>
> var result = $('a');
> var theCount =4; //variable passed in
> var oldCount = -1;
> var theRow = result;
>
> while (theRow.length > 1){
> theRow = jQuery.grep(result, function(n, i){
> return (n && i < theCount && i > oldCount );
> });
> oldCount = theCount - 1;
> theCount = theCount +4;
>
> $(theRow).each(function(i, o){
> // do something to each item of each set});
>
> }
