.ajax（）表单提交错误

Question

I am trying to submit a form using .ajax() jquery function

function submitForm()
{
    $("#quiz").ajaxForm({
               target: '#result',   
              type:   'post',
              beforeSubmit: before_submit,
              success: showResponse

           }).submit();
 }

the problem is form gets submitted in second attempt i.e. when i click twice.

Answer 1

$(document).ready(function(){
    $("form#submit").submit(function() {

    var fname     = $('#fname').attr('value');
    var lname     = $('#lname').attr('value');
        $.ajax({
            type: "POST",
            url: "ajax.php",
            data: "fname="+ fname +"& lname="+ lname,
            success: function(){
                $('form#submit').hide(function(){$('div.success').fadeIn();});

            }
        });
    return false;
    });
});