Convert select into mysql php input - php

Question

I have an order form where the product is selected from a database with the method "select" and "option". I show screenshot and code
Code:
<div class="form-group">
<select class="form-control" name="productName[]" id="productName<?php echo $x; ?>" onchange="getProductData(<?php echo $x; ?>)" >
<option value="">-- Selecciona --</option>
<?php
$productSql = "SELECT * FROM product WHERE active = 1 AND status = 1 AND
quantity != 0";
$productData = $connect->query($productSql);
while($row = $productData->fetch_array()) {
$selected = "";
if($row['product_id'] == $orderItemData['product_id']) {
$selected = "selected";
} else {
$selected = "";
}
echo "<option value='".$row['product_id']."'
id='changeProduct".$row['product_id']."' ".$selected."
>".$row['product_name']."</option>";
} // /while
?>
</select>
</div>
I would like to transform this way of selecting a product (droping-down list using "select") for an order form in a field that can be input the first letters of the name of the product or reference and I can see in the drop-down list those references that match instead of all references in table. I know that "input" and "post" should be used and in the query to the database something like:
// Conexión a la base de datos y seleccion de registros
$con=mysql_connect("localhost","xxxxx_xxxx","xxxxxxx");
$sql = "SELECT * FROM product WHERE referencia, product_name, position_store
like ‘%$buscar%’ ORDER BY id DESC";
mysql_select_db("xxxxxxx_stock", $con);
$result = mysql_query($sql, $con);
The idea is to do what I show in the screenshot:
Once the matches appear, you can select the one that interests you. I'm not sure how to structure the changes to convert the initial code into what I'm looking for.
I appreciate any help.