Hello! On Wed, Mar 25, 2009 at 6:00 PM, Karl Swedberg <k...@englishrules.com> wrote:
> Hi Lay,
> You could do it like this:
> $('table').filter(function() {
> return this.style.tableLayout == 'fixed';
> });
> Not sure what happens when you try it in a browser that doesn't support the
> tableLayout property, though.
Yes, this way works:
$('table').filter(function() {
return this.style.tableLayout=='fixed';
}).css('background-color','#DD0000');
Thank you!
Bye!
Lay
