Relations of cehckbox as submit won't with jquery

Hi Guys,

I want to submit form by checkbox checked with silent mode.
I have the form as following,

<div id="question1">
<form action="<?php echo $submit); ?>" method="post" id="options1" onsubmit="processForm();return true;">
<input type="checkbox" name="agree" value="1" checked="checked" onchange="if(this.checked)  this.processForm() ? $('#confirm_').css('display','block') : $('#confirm_').css('display','none');" id="agree" /><label for="agree">Agree</label>
<?php } else { ?>
<input type="checkbox" name="agree" value="1" onclick="if(this.checked) this.processForm() ? $('#confirm_').css('display','block') : $('#confirm_').css('display','none');" id="agree" /><label for="agree">Agree</label>
<?php } ?>
</form>
</div>

PHP:

Below are the javascript/ajax

<script type="text/javascript">
<!--
$('#confirm').load('index.php?p=form/confirm');
-->
</script>

<script type="text/javascript">
jQuery("#confirm_").click(function() {
   var formData = jQuery("#guest").serialize();

//function processForm() { 
        jQuery.ajax( {
            type: 'POST',
            url: '<?php echo $submit); ?>',
            data: '<?php $data); ?>',
			success: function(data) {
                $('#message').html(data);
            }
        } );
}
</script>

Code (markup):

the issues of these are
1. if checkbox checked that seem no submitted the form to jquery, seems no relations.
2. div section of

<div id="question1">

Code (markup):

does't hiding
3. div section of

<div id="confirm_" style="<?php echo ($agree) ? 'display:block;' : 'display:none;' ?>">
<div id="confirm"></div>
</div>

PHP:

not show...

I really needed some helps and pointers form You Guys, I really appreciate every suggestions and many thanks.

 

You honestly got this javascript idea all wrong...

First bug :
onsubmit="processForm();return true;"
That means that it will submit the form , so it isn't silent anymore...

2nd bug , from what I see in your code :
<?php $data); ?>

actually 3 bugs here :
1.Either echo / = $data;
2.remove the ) before the ;
3.You did not mention the name of the post variable.Meaning that if you read the variable in php as $_POST['variable_name'] in your ajax request the data section should look like :
data : { 'variable_name' : variable_value }
where usually variable_values is a javascript variable , meaning that the data you want to send should be stored like :

var variable_value = "'"+<?php echo $data; ?>+"'";

If you would have posted the full code I would probably be able to help more...

 

Hi tvoodoo,
Thanks for your Care and corrections, I appreciate it. but this is already solved

 

Well then post and tell people that are trying to help you that it is solved and the solution that worked for you , maybe other people need it as well and want to learn from it....