Dynamic Dependent Dropdown - php

Question

I have a form that pulls some dropdown data from an existing db. I've been working on a second dropdown that references the first to get more specific information from a different DB, however it looks like my code is broken somewhere. The first dropdown is populated fine but when i choose a "Manager" the Site dropdown goes blank, I even lose the "Select Site" option.
Any help would be appreciated.
<script src="https://code.jquery.com/jquery-2.1.1.min.js" type="text/javascript"></script>
<script>
function getSite(val) {
$.ajax({
type: "POST",
url:"get_site.php",
data:'manager_id='+val,
success: function(data){
$("#site-list").html(data);
}
});
}
</script>
html/php
Manager<br/>
<select name="manager_id" onChange="getSite(this.value);">
<option value="">Select Manager</option>
<?php
$results = mysql_query("SELECT * FROM _managers");
while ($row_unit = mysql_fetch_array($results)){
?>
<option value="<?php echo $row_unit["id"]; ?>"><?php echo $row_unit["company"]; ?></option>
<?php
}
?>
</select>
<br/><br/>
Site<br/>
<select name="site_id" id="site-list">
<option value="">Select Site</option>
</select>
get_site.php
<?php
include('includes/connect-db.php');
if(!empty($_POST["manager_id"])) {
$manager_id = $_POST["manager_id"];
$results = mysql_query("SELECT * FROM _sites WHERE manager_id = $manager_id");
?>
<option value="">Select Site</option>
<?php
while ($row_site = mysql_fetch_array($results)){
?>
<option value="<?php echo $row_site["id"]; ?>"><?php echo $row_site["site_name"]; ?></option>
<?php
}
}
?>

Answer 1

As per discussion in comment.
I made the adjustment but still not getting my values from the
"get_site.php" file. Although now the "Select Site" stays in the site
dropdown.
Assuming you are getting proper data from MySQL server do some changes in get_site.php as below.
get_site.php
<?
include 'includes/connect-db.php';
if ((!empty($_POST["manager_id"])) && (isset($_POST["manager_id"])))
{
$manager_id = $_POST["manager_id"];
$results = mysql_query("SELECT * FROM _sites WHERE manager_id = '{$manager_id}'");
$options = "<option value=''>Select Site</option>";
while ($row_site = mysql_fetch_assoc($results))
{
$options .= "<option value='{$row_site['id']}''>{$row_site['site_name']}</option>";
}
return $options; // I personally prefer to echo using json_encode and decode it in jQuery
}
?>
Above code should give you the data you want.
Hope this solves your issue.Do comment if you are having any difficulties.