Skip hidden inputs in Jquery serialize

Question

I have forms with inputs of hidden and visible like below.In this example I want to serialize only name=\'country\' which is not hidden

Answer 1

You can make a temporary copy of it and remove the hidden inputs from it before serializing :

var form = $("#finalform").clone();
$(form).find("input[type=hidden]").remove()
var serialized = $(form).serializeArray();

Answer 2

Here is another possible solution with a demo.

var notHidden = $('#finalform > :not(input[type=hidden])').serializeArray();

$('#itemsSerializedCount').text($(notHidden).length)

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/json2/20140204/json2.js"></script>
<form id="finalform">
<input type='hidden' name='sid' value='901271316' />
<input type='hidden' name='tco_currency' value='USD'>
<input   name='country' value='India' />
</form>

<lablel id="itemsSerializedCount"></label>

Here, the input control which is not hidden is serialized.

Answer 3

I find this a cleaner way of doing it:

var frm = $('#finalform [type!="hidden"]').serialize()

Answer 4

use below code . not(:hidden) will not include input type hidden in serialize()

check DEMO

var frm  = $('#finalform').find(":input:not(:hidden)").serialize();

Or

var frm  = $('#finalform :input:not(:hidden)').serialize();

DEMO

var frm  = $('#finalform :input:not(:hidden)').serializeArray();

var frm1  = $('#finalform').find(":input:not(:hidden)").serializeArray();

$('#firstOutput').text(JSON.stringify(frm))
$('#secondOutput').text(JSON.stringify(frm1))

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.0/jquery.min.js"></script>
<form id="finalform">
 <input type='hidden' name='sid' value='901271316' />
 <input type='hidden' name='tco_currency' value='USD'>
 <input   name='country' value='India' />
</form>


<div id="firstOutput"> </div>
<div id="secondOutput"> </div>