PHP - Splitting For Dropdown Into 2 Variable

I have been pulling my hair out for about 2 weeks with this problem.  I am doing this on a localhost set up, not a live server.  Here are the fields in the categories table in question:

id, cattitle, caturl, catdesc, orderby, parentid

What I am trying to accomplish is to create a dropdown list from the values in cattitle and parentid fields which I have done successfully.
Here is the code from my "categories_modify.php" file where the dropdown is displayed. 

<form action="" method="POST">

<table cellspacing="5" cellpadding="1" border="0">
<tr>
<td>
Change Parent Category To:
</td>
<td>
<select name="location">
<?php foreach ($categories as $loc): ?>
<option name="parid" value="<?php htmlout($loc['parentid']); ?>"><?php htmlout($loc['cattitle']); ?></option>
<?php endforeach; ?>
</select>
</td>
</tr>
<tr>
<td>
</td>
<td>
<input type="hidden" name="id" value="<?php htmlout($id); ?>" />
<input type="submit" value="Update" />
</td>
</tr>
</table>
</form>

The code generating the values in the form is in my controller script called index.php.  Here is the code for that part.  **Note that the top section of the script below populates a series of text boxes with current values in the database for editing/updating I did not show "edit" form in the code above as it works fine.  The part of code populating my dropdown box values starts with the "locat... line except for my db.inc.php database connection at the top (which also works fine).

if (isset($_POST['action']) and $_POST['action'] == 'Edit')
{
include $_SERVER['DOCUMENT_ROOT'] . './includes/db.inc.php';
  {   
   $id = mysqli_real_escape_string($link, $_POST['id']);
   $sql = "Select id, cattitle, orderby, caturl, catdesc, parentid from categories where id = '$id'";
   $result = mysqli_query($link, $sql);
   if (!$result)
   {
   $error = 'Error fetching category details ' . mysqli_error($link);
   include 'error.php';
   exit();
   }

   $row = mysqli_fetch_array($result);
   $pagetitle = 'Edit Category';
   $action = 'editform';
   $id =  $row['id'];
   $cattitle = $row['cattitle'];
   $orderby =  $row['orderby'];
   $caturl =  $row['caturl'];
   $catdesc =  $row['catdesc'];
   $button = 'Update';
   
   $locate = "Select id, cattitle, parentid from categories order by cattitle asc";
   $res = mysqli_query($link, $locate);
   if (!$res)
   {
   $error = 'Error fetching category details ' . mysqli_error($link);
   include 'error.php';
   exit();
   }
      
   while ($locat = mysqli_fetch_array($res))
   {
   $categories[] = array('id' => $locat['id'], 'cattitle' => $locat['cattitle'], 'parentid' => $locat['parentid']);
   }

 }
   include 'categories_modify.php';
   exit();
}

Now for the code that gets the POSTED info from the dropdown window. 

if (isset($_GET['location']))
{
   include $_SERVER['DOCUMENT_ROOT'] . './includes/db.inc.php';
   
   $id = mysqli_real_escape_string($link, $_POST['id']);
   $parentid = mysqli_real_escape_string($link, $_POST['parid']);

   $parentid++;

   $sql = "Update categories set
      parentid = '$parentid'
      where id = '$id'";

   if (!mysqli_query($link, $sql))
   {
   $error = 'Error updating categories ' . mysqli_error($link);
   include 'error.php';
   exit();
   }

   header('Location: .');
   exit();
}

The goal is to take the parentid value and increase it by 1 so I can create subcategories based on the parentid value.  (maybe not the best way but I'd still like to figure out why I can't make this work.)  This line is where I think the problem is, or atleast part of it:

   $parentid = mysqli_real_escape_string($link, $_POST['parid']);

I believe it is coming back blank because when I re-run the script for my "test" value it will always comeback as 1 (the default value for parentid is 0) 0 + 1 = 1.  If I update the form again where the parentid now equals 1 it still equals 1, can't get it to 2.
BTW I have tried every variation of the above line I could think of with no luck hence this post.
Thanks for any help.  I hope this was clear enough.

Hopefully someone can help, firstly i am a newbie at php coding but keen to learn more.   Here is my problem    The database is built correctly with all the information i need, the first part is the dropdown menu which you select a car, next you choose a month and mileage - from there it shall show you the correct price for each one. There is columns for each price, mileage and months i just need to figure out how best to start this thing.   I first started by this   $query = "SELECT DISTINCT Full_Description FROM import"; 

Hi all, here's my code:

Code: [Select]
<?php
foreach ($_SESSION['topping'] as $value) {
    echo "<tr><td width='30%'>Topping</td><td width='50%'>$value</td><td width='20%'><select name='notopping'>";
foreach ($_SESSION['cupcake'] as $number) { '<option name="notoppings[]" value="'.$number.'">".$number."</option>'; }
echo "</select></td></tr>";
}
?>

$_SESSION['cupcake'] is a value from either 6, 12, 24 or 36.  What I want to do is put them into a drop down box (second foreach) as the value and the displayed value - counting up from 1 (so 1,2,3,4,5,6 or up to 12,24 etc).

Also by creating this as an array, does this mean than for each topping (say Vanilla and Chocolate) the value dynamically created can be used on the next page by using $_POST['notoppings'] to display each type (two different numbers - one for Vanilla and one for Chocolate).  Does that make sense?

Thanks!

Jason

I have found postings close, but not close enough to find my error.
I am looking up data from a MySql table and putting it in a dropdown box on a form. I can select the item, but apparently not really. I am not able to echo it, or post it to a record.
I'm sure I am missing something simple, but...
Code attached if anyone can show me the errors of my ways.
Thank you.

I have a dropdown list which have some elements.
I want to get all the elements of the dropdown list into a variable of type array.

Any Idea???