Pass or Post the input value without using form - php

Question

I'm working right now on how I can pass the value of <input id="sub" name="sub"> that has a value of first 3 letters of dropdown option text. I need to pass or post the value of <input id="sub" name="sub"> to process my query below.
When I choose for example the ITEquipment in dropdown, the <input name="sub"> get the first 3 letters of dropdown option text. So now it has the value of ITE, that ITE what I need in my query. In query for example, $sub=$POST["sub"] is equal to ITE so I used LIKE function, LIKE '$sub-__' . Hope you understand what I'm trying to do.
How I can do that without using submit form?
Here's my example process of what I'm working http://jsfiddle.net/xqGLS/1/
Help please?
<?php
$mysqli = new mysqli("localhost", "root", "", "2015");
$resultcode = $mysqli->query("SELECT category, id, maincode FROM category GROUP BY id ORDER BY maincode");
$code = '';
while($row = $resultcode->fetch_assoc())
{
$code .= '<option value = "'.$row['maincode'].'">'.$row['category'].'</option>';
}
?>
<form action="<?php $_SERVER['PHP_SELF']?>" method="POST">
<br/>
Category
<select name="maincode" style="text-transform:uppercase;" onchange = "GetChangedValue(this);">
<option value=""></option>
<?php echo $code; ?>
</select>
</br>
Sub Code
<input type="" name="sub1code" id="sub1code" value="" readonly style="width:45px;text-transform:uppercase;">
<script>
$('[name="maincode"]').change(function() {
$('[name="sub1code"]').val($(this).val());
var input = $('[name="sub"],[name="sub1"]'),
input1 = $('[name="sub"]'),
input2 = $('[name="sub1"]'),
input3 = $('[name="equal"]');
input.change(function () {
input3.val(input1.val() + input2.val());
});
});
</script>
<script>
function GetChangedValue(e) {
var value = e.options[e.selectedIndex].text;
var elem = document.getElementById("sub"); elem.value = value.substring(0,3);
}
</script>
<input name="sub" id="sub" value="" style="width:35px;text-transform:uppercase;" readonly>
--My Query--
<input id="sub1" name="sub1" style="width:35px;text-transform:uppercase;" value='
<?php
$result2 = $mysqli->query("SELECT * FROM code
WHERE sub1code LIKE 'SUP-___' ORDER BY sub1code");
while($row = $result2->fetch_assoc())
{
$value = $row['sub1code'];
}
$first = substr($value, 0, 4);
echo $first;
$last = substr($value, -3);
$i="0";
while($i<=$last)
{
$i++;
}
$value2=strlen($i);
echo $first;
if($value2==1)
{
echo "00".$i;
}
elseif($value2==2)
{
echo "0".$i;
}
else
{
echo $i;
}
?>'>
<input id="equal" name="equal" value="" style="width:60px;text-transform:uppercase;" type="hidden">
<input type="submit" name="">
</form>

Answer 1

HEAVILY EDITED
$(document).ready(function(){
$("#dd").change(function(){
var v1 = $(this).val().substring(0,3);
$.ajax({
type : 'get',
url : 'call.php',
data : {varReq:v1}
}).done(function(returnResult){
alert(returnResult);
});
});
});
call.php would contain your php/mysql script. To get the value passed through there should be via $_GET or $_POST. But you have to include the proper type in your jquery's ajax too.
So in my example, What I have on my php, just a dummy one.
<?php
if(isset($_GET['varReq']))
{
if($_GET['varReq'] == 'ABC') echo "AAAA";
else echo "BBBB";
}
?>
It pass this html element...
<h1>Example</h1><p>BBBB</p>
Then pop up message (alert) will be called showing the return value.