I wouldn't use either version.
Instead, I would change your CSS from:
tr.rowodd { background-color: #FFF; }
tr.roweven { background-color: #F2F2F2; }
to:
tr { background-color: #FFF; }
tr.roweven { background-color: #F2F2F2; }
and then use just one line of jQuery code:
$('#foobar tr:visible:odd').addClass('roweven');
Now you're doing only half the work and letting the CSS cacading rules take
care of the rest.
-Mike
On Fri, Jan 1, 2010 at 10:38 AM, Paul Kim <[email protected]> wrote:
> Thanks for your reply. Your solution works. I had a feeling that :even and
> :odd filters are zero-based, but found that to be "odd" in this situation.
> So now that I have 2 ways to stripe visible table rows using jQuery, which
> solution do you prefer?
>
> $('#foobar tbody tr:visible:even').addClass('rowodd');
> $('#foobar tbody tr:visible:odd').addClass('roweven');
>
> or
>
> $('#foobar tbody tr:visible').each(function(i) {
> if ((i+1) % 2 === 0) {
> $(this).addClass('roweven');
> }
> else {
> $(this).addClass('rowodd');
> }
> });
>
> I guess both solutions work so it really doesn't matter, but which method
> would you choose? The first solution contains less code but the second
> solution seems more intuitive.
>
>
>
> 2010/1/1 Šime Vidas <[email protected]>
>
> Also, you really don't need two counters (i and j)....
>>
>> var rows = $('#foobar tbody tr:visible');
>> for (var i = 0; i < rows.length; i++){
>> if ((i + 1) % 2 == 0) {
>> rows.eq(i).addClass('roweven');
>> }
>> else {
>> rows.eq(i).addClass('rowodd');
>> }
>> }
>>
>> However, don't use the for loop, you have jQuery's each method...
>>
>> $('#foobar tbody tr:visible').each(function(i) {
>> if ((i+1) % 2 === 0) {
>> $(this).addClass('roweven');
>> }
>> else {
>> $(this).addClass('rowodd');
>> }
>> });
>>
>
>
