PHP - Drop Down Menu Retrieve Data Help

foreach ($_SESSION["products"] as $cart_itm)
        {
           $product_code = $cart_itm["code"];
		   $results = $mysqli->query("SELECT * FROM products WHERE product_code='$product_code' LIMIT 1");
		   $obj = $results->fetch_object(); 

I have some php code on view invoice and it's getting all the data I need to from three database tables but for some reason it's not getting the data for one specific column and unsure why

I have done a echo sql and it is echoing the sql query ok and have ran the sql query in phpmyadmin and is displaying the column in phpmyadmin that is not displaying on the php page so am bit confused, below is the coding I have. I took out a lot of it and tried to keep to the relevant parts

<?php
$sql = "SELECT t1.id, t1.invoice_number, DATE_FORMAT(t1.invoice_date,'%d/%m/%Y') AS invoice_date, t1.user_id, t2.invoice_number, t2.service_name, t2.service_price, t3.user_id, t3.customer_name, t3.customer_phone, t3.customer_email, t2.service_name, t2.service_price, t1.invoice_total, t1.balance, t1.notes, DATE_FORMAT(t1.payment_date,'%d/%m/%Y') AS payment_date, t1.payment_method
FROM invoices as t1
LEFT JOIN invoice_products as t2
ON t1.id = t2.invoice_id
LEFT JOIN users as t3
ON t1.user_id = t3.user_id
WHERE t1.user_id = '".$_SESSION['user_id']."' and t1.id = '".$_GET['id']."'";

$result = mysqli_query($connect, $sql);

                if(mysqli_num_rows($result) > 0)
                {
                if($row = mysqli_fetch_array($result))  
                {
                ?>

                Notes: <?php echo $row['notes'];?>

                <?php
                    }
                    }
                    ?>

I am trying to create a code where the user enters the zip code and if found in the database, a location will be retrieved. I am getting two undefined variables as errors. I am still learning the POST and GET method and I fear that is where I am getting something mixed up. I have also attached an image of my DB.

Thank you so much any input is greatly appreciated.

Notice: Undefined variable: address in C:\xampp\htdocs\index.php on line 53

Notice: Undefined variable: zip in C:\xampp\htdocs\index.php on line 54

index.php file code:

<?php

     include ('header.php');
     include ('function.php');

     ?>


        <div class="wrapper">
           <div>
               <?php
               echo'<form method="POST" action"'.getLocations($conn).'">
                   <input type="text" name="zip" class="search" placeholder="Zip code"><br>
                   <button type="submit" value="submit" id="submit">Submit</button>
                   </form>';
               getLocations($conn);
              ?>         
           </div>
        </div>
        
        <div>
             <!--foreach($zip as $zip) : -->
            <?php 
                  echo $address['address'];
                  echo $zip['zip']; 
            ?>
        </div>
    
    
 function.php file code:
    
    
    <?php
    
    $dBServername = "localhost";
    $dBUsername = "root";
    $dBPassword = "";
    $dBName = "addresses";
    
    // Create connection
    $conn = mysqli_connect($dBServername, $dBUsername, $dBPassword, $dBName);
    
    // Check connection
    if (!$conn) {
        die("Connection failed: " . mysqli_connect_error());
    }
    function getLocations($conn) {
        if (isset($_POST['submit'])){
            
             // Validate Zip code field
            if (!empty ($_POST['zip']) && is_numeric ($_POST['zip'])) {
                
            $zip = (int)$_POST['zip'];
        
            
        $query = "SELECT * FROM locations WHERE zip = '$zip'";
        
        //get the results
        $result = mysqli_query($conn, $query);
        
        //fetch the data
        $zip = mysqli_fetch_all($result, MYSQLI_ASSOC);
        var_dump($zip);
        
        
        mysqli_free_result($result);
        
        //close connection
        mysqli_close($conn);
    
        
       }
    } 
      
    }