I am kinda confused on it, when trying to send value on the same page.
You should wrap your code with
$(document).ready(function(){
// your code here
});
This way, it will only run when the browser finishes processing the structure of your HTML.
UPDATE
There was a lot of debug stuff on your code, try this (requires Firebug to see the output of the ajax request):
<script>
$(document).ready(function(){
$("select[name='sweets']").change(function () {
jQuery.ajax({
type: "POST",
data: $("form#a").serialize(),
success: function(data) {
// Check the output of ajax call on firebug console
console.log(data);
}
});
});
});
</script>
Here is the working code for you. To send ajax request to the same page you can keep url
parameter empty, which you are already doing. If you are trying to make the script behave differently when $_POST
has value then use isset
as I have used below.
<?php
if(isset($_POST['sweets']))
{
echo $_POST['sweets'];
exit;
}
?>
<script>
$(function(){
$("select[name='sweets']").change(function () {
var str = "";
$("select[name='sweets'] option:selected").each(function () {
str += $(this).text() + " ";
});
jQuery.ajax({
type: "POST",
data: $("form#a").serialize(),
success: function(data){
jQuery(".res").html(data);
$('#test').html(data);
}
});
var str = $("form").serialize();
$(".res").text(str);
});
});
</script>
<div id="test">
</div>
<form id="a" action="" method="post">
<select name="sweets" >
<option>Chocolate</option>
<option selected="selected">Candy</option>
<option>Taffy</option>
<option>Caramel</option>
<option>Fudge</option>
<option>Cookie</option>
</select>
</form>