jQuery驗證插件1.9.0 ..我希望提交在有效時不會重新加載頁面

Question

$("#da-ex-validate2").validate({
                rules: {
                    details: {
                        required: true,
                        rangelength: [1, 500]
                    },
                    editor1: {
                        required: true,
                        minlength: 1
                    },
                    title: {
                        required: true,
                        rangelength: [1, 100]
                    },
                    SlideDeckPhoto: {
                        required: "#iButton:checked",
                        accept: ['.jpeg', '.png', '.jpg', '.gif']
                    },
                    min1: {
                        required: true,
                        digits: true,
                        min: 5
                    },
                    max1: {
                        required: true,
                        digits: true,
                        max: 5
                    },                          
                    submitHandler: function(form) {
                        $(form).ajaxSubmit();
                    },
                    range1: {
                        required: true,
                        digits: true,
                        range: [5, 10]
                    }

                },
                invalidHandler: function (form, validator) {
                    var errors = validator.numberOfInvalids();
                    if (errors) {
                        var message = errors == 1
                        ? 'You missed 1 field. It has been highlighted'
                        : 'You missed ' + errors + ' fields. They have been highlighted';
                        $("#da-ex-val2-error").html(message).show();
                    } else {
                        $("#da-ex-val2-error").hide(); // it's not work !!! and the page is reload !!
                    }
                }
            });

我也想將表單值保存到MySql而無需重新加載頁面。 請幫忙 ！ 我讀了很多文章，嘗試了很多事情！ 如果您輸入代碼，請告訴我們在哪里放置。.順便說一句，我的表單有幾個輸入字段，還有一個文件輸入字段。 請幫忙 ！

Answer 1

從提交處理程序返回false

submitHandler: function(form) {
    $(form).ajaxSubmit();
    return false;
}

Answer 2

這是我的解決方案，將其放在invalidHandler之后：

            submitHandler: function(form) {
                var dataString = $('#YourFormID').serialize();
                $.ajax({
                    type: 'POST',
                    url: 'yourpage.php',
                    data: dataString,
                    success:  function() {
                        $('#YourErrorDivID').hide();
                        $('#YourSuccessDivID').html("Your Message").show();
                    }
                });
            }