Ajax call to same page is not working - php

Question

I am getting records from database in WordPress then creating and adding value in select tag(HTML) dynamically.
<?php
global $wpdb;
$registeredUsers = $wpdb->get_results('SELECT * FROM wp_users where user_login != "Admin"',ARRAY_A);
$select='<select name="users" class="form-control" id="users">';
$select.= '<option value="Select User"> Select User</option>';
foreach($registeredUsers as $user)
{
$select.='<option value="'.$user['user_email'].'">'.$user['user_login'].'</option>';
}
$select.='</select>';
?>
I am using $select variable in Html and drop down is being displayed properly.
<form id="a" action="" method="post">
<div style="margin: 0 auto;width:500px;">
<?php echo $select ?>
</div>
</form>
I have written code to get selected drop down onchange event in jquery. It return success but I am not able to get selected value of dropdown.
<script type="text/javascript">
$(document).ready(function(){
$("select[name='users']").change(function () {
jQuery.ajax({
type: "POST",
data: $("form#a").serialize(),
success: function(data){
alert("SUCCESS");
}
});
});
});
</script>
Below code return nothing
if(isset($_POST['users'])) {
echo $_POST['users'];
}

Answer 1

Set url option to your page in your ajax function:)
<script type="text/javascript">
$(document).ready(function(){
$("select[name='users']").change(function () {
jQuery.ajax({
type: "POST",
url:"yourpage.php"
data: $("form#a").serialize(),
success: function(data){
alert("SUCCESS");
}
});
});
});
</script>
replace yourpage.php