在null中调用成员函数prepare（）

Question

我需要报告以调用null中的成员函数prepare（）的报告

$servername = "localhost";
$username = "xxxxxxxxx";
$password = "xxxxxxxxx";
$dbname = "xxxxxxxxx";
$connect_pdf = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password);
function fetch_customer_data($connect_pdf)
{
    $query_1 = "SELECT ad,adet,urunfiyat FROM urunler b LEFT JOIN bilgi c ON  (c.id=b.urun) where b.ust='".$_GET["id"]."'";
    $query_2 = "SELECT id,ad,soyad,adres,fiyat,email FROM siparisler WHERE id='$_GET[id]'";
    $statement = $connect_pdf->prepare($query_1);
    $statement->execute();
    $result_1 = $statement->fetchAll();

在null中调用成员函数prepare（）

$output = '
<div class="table-responsive">
    <table class="table table-striped table-bordered">
        <tr>
            <th>Name</th>
        </tr>
';

我需要报告以调用null中的成员函数prepare（）的报告

foreach($result_1 as $row)
{
    $output .= '
        <tr>
            <td>'.$row["ad"].'</td>
        </tr>
    ';
}
$output .= '
    </table>
</div>
';
return $output;
}