problem with slide and ajax

i have this structure in my html

 <div id ="slide1">
            <form method="post" id="customForm" action="">
                //my content - name, email, mypassword, pass2
            </form> //i already tried delete this
        </div>
        <div id ="slide2">
            <form method="post" id="customForm1" action="">  //and this, to have only one form obviously this didn't work, because the div of slide1 is already closed.
                //my content - other stuff
            </form>
        </div>

and i have this JS

 $(document).ready(function() {
        $("#customForm").submit(function() {
            $.ajax({
                url: "validation1.php",
                type: "post",
                dataType: "json",
                data: {
                    name: $('#name').val(),
                    email: $('#email').val(),
                    myPassword: $('#myPassword').val(),
         开发者_开发技巧           pass2: $('#pass2').val()
                },
                success: function(data) {
//stuff
}

the problem is. I want to group the data of tho slides and send to validation1.php, but for that i must have a only form.

If the problem is not clear please tell me.

thanks

---

If I understand correctly you are saying that you need to send the 2 slides in one post but in your html the data lives in 2 forms elements.

You can always build a new form with the data you want in it and just submit that, but also review the use of forms on your page anyway.

var href = '.... where ever you are sending the data .....';
var form = $('<form method="post" action="' + href + '"></form>');

var d1=$('.slide1 some subdata selector');
var d2=$('.slide2 some subdata selector')

$('<input type="hidden" name="data1">').val(d1).appendTo(form);
$('<input type="hidden" name="data2">').val(d2).appendTo(form);
$('body').append(form);
form.submit();
form.detach();