updating SQL query from a dropdown - php

Question

I've got a dropdown menu that is supposed to update my SQL query. If I create the dropdown from the select HTML, it works fine, but when I run it from a PHP array it doesn't select the other options when posted (though it does select them if I manually select the array element from in the program.
Moreover, I've put several echo statements throughout the program to trouble shoot, and I can clearly see that the correct item is being selected, AND that it's posting to the SQL code, but it's just not working right.
For example, my echo statements print out the SQL query
sql = SELECT * from finished_goods WHERE BrandDesc LIKE '%Alma Rosa%'
which works fine. but if I use the drop down to select "alere" it doesn't work.
sql = SELECT * from finished_goods WHERE BrandDesc LIKE '%Alere%'
<table>
<form id = "myform" method="post" action="">
<th>FG id : <input type = "text" name = "finished_goods_id2"/></th>
<th>Product No : <input type = "text" name = "ProdNo"/></th>
<th>Product Name : <input type = "text" name = "ProductName"/></th>
<th>Product Group : <input type = "text" name = "ProductGroup"/></th>
<!-- this code puts a drop down in, but it doesnt update the form -->
<th>Brand : <!-- this select statement works fine
<select name = "BrandDesc">
<option value = "Alma Rosa">Alma Rosa text</option>
<option value = "Alere">Alere</option>
<option value = "Hitching Post">Hitching Post text</option>
</select>
-->
<?php // this array replaces the select dropdown, but doesn't update
$name = 'BrandDesc';
$selected = 0;
$options = array( 'Alma Rosa', 'Alere', 'Hitching Post' );
echo dropdown( $name, $options, $selected );
?>
<th>Varietal : <input type = "text" name = "Varietal"/></th>
<th>Vintage : <input type = "text" name = "VintYear"/></th>
<th>Quantity :</th>
<th><input type = "submit" name = "send" value = "Submit"/></th>
<tr></tr>
<?php
// new connection
$conn = dbConnect('read', 'thereal8_work', 'PDO'); // database for online
$brandNo = 0; // sets the brand so that it displays the first select statement
$brandAccess = $options[$brandNo];
echo " first Brand Access = ".$brandAccess. " <br>";
if ($_POST['BrandDesc']) {
echo " base brandNo = ".$brandNo. "<br>";
$brandNo = $_POST['BrandDesc'];
echo " posted brandNo = ".$brandNo. " <br> ";
echo "options in the function = ".$options[$brandNo]. "<br>";
//$brandAccess = $_POST['BrandDesc'];
$brandAccess = $options[$brandNo];
echo "submitted brand = ".$brandAccess. " <br> ";
}
$sql = "SELECT * from finished_goods WHERE BrandDesc LIKE '%$brandAccess%' ";
echo "sql = ".$sql." <br>";
// get the result
Thanks,