从两个依赖的下拉列表中获取表单数据到php

我的页面上有一个表单，其中包括2个依赖的下拉列表。当用户从第一个列表中选择值时，它将填充第二个列表，然后用户从第二个列表中选择值。

我想提交表单数据到php页面插入到表在mysql中，但当它提交，所有的数据是通过除了值从第二列表。第一个列表和其他输入字段的值传递OK。我已经试过了所有我知道的方法，但还是不行。有什么想法如何实现这一点吗?

这是来自index2.php的表单(编辑:简化了form元素):

<form name="part_add" method="post" action="../includes/insertpart.php" id="part_add">
                <label for="parts">Choose part</label>
                    <select name="part_cat" id="part_cat">
                        <?php while($row = mysqli_fetch_array($query_parts)):?>
                        <option value="<?php echo $row['part_id'];?>">
                        <?php echo $row['part_name'];?>
                        </option>
                        <?php endwhile;?>
                    </select>
                    <br/>
                <label>P/N</label>
                    <select name="pn_cat" id="pn_cat"></select>
                <br/>
                <input type="text" id="manufactured" name="manufactured" value="" placeholder="Manufactured" />
                <input id="submit_data" type="submit" name="submit_data" value="Submit" />
            </form>

javascript:

$(document).ready(function() {
$("#part_cat").change(function() {
    $(this).after('<div id="loader"><img src="img/loading.gif" alt="loading part number" /></div>');
    $.get('../includes/loadpn.php?part_cat=' + $(this).val(), function(data) {
        $("#pn_cat").html(data);
        $('#loader').slideUp(200, function() {
            $(this).remove();
        });
    });
});
});

这是php加载第二个列表:

<?php 
include('db_connect.php');
// connects to db
$con=mysqli_connect(DB_HOST,DB_USER,DB_PASS,DB_NAME);
$part_cat = $_GET['part_cat'];
$query = mysqli_query($con, "SELECT * FROM pn WHERE pn_categoryID = {$part_cat}");
while($row = mysqli_fetch_array($query)) {
echo "<option value='$row[part_id]'>$row[pn_name]</option>";
}
?>

我得到$part_cat从第一个列表insertpart.php，但$pn_cat.

编辑:这是insertpart.php(简化，它只是回显结果)

<?php
//Start session
session_start();
//Include database connection details
require_once('../includes/db_details.php');
//DB connect
$con=mysqli_connect(DB_HOST,DB_USER,DB_PASS,DB_NAME);
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
// escape variables for security
// find part name based on ID
$part_typeID = mysqli_real_escape_string($con, $_POST['part_cat']);
$part_name_result = mysqli_query($con, "SELECT part_name FROM parts WHERE part_id = $part_typeID");
$part_row = mysqli_fetch_array($part_name_result, MYSQL_NUM);
$part_type = $part_row[0];
echo"part_type='$part_type'";
//find pn value based on id
$pn_typeID = mysqli_real_escape_string($con, $_GET['pn_cat']);
$pn_name_result = mysqli_query($con, "SELECT pn_name FROM pn WHERE pn_id = $pn_typeID");
$pn_row = mysqli_fetch_array($pn_name_result, MYSQL_NUM);
$pn = $pn_row[0];
echo"pn='$pn'";
mysqli_close($con);
?>

它仍在进行中，所以代码是丑陋的，我知道我混合POST和GET正在纠正。如果我在此页上返回$pn_cat，则没有输出，$part_type是OK的。