Question

Possible Duplicate:
don't work 'fadeOut' If there is 'replaceWith'.?

I want after delete row(s) in table, for it row that deleted use of effect fadeOute('slow'), but it don’t work in my code.(if not use of .replaceWith() it worked.)

EXAMPLE => checked and clicke on DELETE

code:

$.ajax({
        type: "POST",
        url: url,
        data: dataString,
        cache: false,
        success: function(html){
            var $html = $(html);
////////////////////////////////// here ////////////////////////////////
                    $('.ser_form #paginate input:checkbox:checked').parent().parent().fadeOut("slow");

                    $('#num_count').replaceWith($html.find('#num_count'));
                    $('tr#paginate').replaceWith($html.find('tr#paginate'));
                    $('.pagination').replaceWith($html.find('.pagination'));
////////////////////////////////// here ////////////////////////////////

        },

Answer 1

Do the fadeOut first, than when it’s completed, delete the element

element.fadeOut("slow", function(){
    $(this).remove();
});