You need to use ajaxForm() for that functionality. ajaxSubmit() will not do it.
Mike On 1/15/07, yi huang <[EMAIL PROTECTED]> wrote: > I've found a similar question here > (http://jquery.com/pipermail/discuss_jquery.com/2006-August/009469.html) > . > But it seems that it's still not resolved . > > -- > http://codeplayer.blogspot.com/ > _______________________________________________ > jQuery mailing list > discuss@jquery.com > http://jquery.com/discuss/ > > > _______________________________________________ jQuery mailing list discuss@jquery.com http://jquery.com/discuss/
