Получение значений SQL с PHP в jsPDF

В настоящее время я работаю над системой счетов и пытаюсь экспортировать счета в формате PDF. Создание PDF должно быть выполнено с помощью jsPDF, а значения для счета-фактуры (Customer, Adress, Ordernumber и т. Д.) Находятся в базе данных SQL.
Мой план состоял в том, что программа .js, которая создает PDF-файл, получает документ из другого PHP-документа.

var AnRD =      <?php include "GetData.php"; return json_encode(giveAnRD(ResID), JSON_HEX_TAG);?>;;
var AbRD =      <?php include "GetData.php"; return json_encode(giveAbRD(ResID), JSON_HEX_TAG);?>;;
var VName =     <?php include "GetData.php"; return json_encode(giveVorname(ResID), JSON_HEX_TAG);?>;;
var NName =     <?php include "GetData.php"; return json_encode(giveNachname(ResID), JSON_HEX_TAG);?>;;
var Strasse =   <?php include "GetData.php"; return json_encode(giveStarsse(ResID), JSON_HEX_TAG);?>;;
var Hnummer =   <?php include "GetData.php"; return json_encode(giveHnummer(ResID), JSON_HEX_TAG);?>;;
var PLZ =       <?php include "GetData.php"; return json_encode(givePLZ(ResID), JSON_HEX_TAG);?>;;
var Ort =       <?php include "GetData.php"; return json_encode(giveOrt(ResID), JSON_HEX_TAG);?>;;
var Anrede =    <?php include "GetData.php"; return json_encode(giveAn(ResID), JSON_HEX_TAG);?>;;

Документ GetData.php выглядит так:

include "../dblogin.php";
$resID="";$db = new PDO("mysql:host=$dbhost;dbname=$dbname;", "$dbuser", "$dbpass");
//##############################################
function giveAnRD($resID)
{
$AnRD = "SELECT reserviert_von FROM reservierungen WHERE ReservierungsID = $resID";
return $AnRD;
}
//##############################################
function giveHausName($resID)
{
$HID = "SELECT HausID FROM reservierungen WHERE ReservierungsID = $resID";
$AnRD = "SELECT Hausname FROM hauser WHERE HausID = $HID";
return $AnRD;
}
//##############################################
function giveAbRD($resID)
{
$AnRD = "SELECT reserviert_bis FROM reservierungen WHERE ReservierungsID = $resID";
return $AnRD;
}
//##############################################
function giveAnrede($resID)
{
$PID = "SELECT PersonID FROM reservierungen WHERE ReservierungsID = $resID";
$TITEL = "SELECT Titel FROM persons WHERE PersonID AND IS NOT NULL = $PID";

if ($TITEL =! NULL)
{
$Anrede = $GEN . ' ' . $TITEL;
}

else
{
$Anrede = $GEN;
}
$GEN = "SELECT Anrede FROM persons WHERE PersonID = $PID";
return $GEN;
}
//##############################################
function giveVorname($resID)
{
$PID = "SELECT PersonID FROM reservierungen WHERE ReservierungsID = $resID";
$TITEL = "SELECT Titel FROM persons WHERE PersonID AND IS NOT NULL = $PID";
return $TITEL;
}
//##############################################
function giveNachname($resID)
{
$PID = "SELECT PersonID FROM reservierungen WHERE ReservierungsID = $resID";
$TITEL = "SELECT Titel FROM persons WHERE PersonID AND IS NOT NULL = $PID";
return $TITEL;
}
//##############################################
function giveStrasse($resID)
{
$PID = "SELECT PersonID FROM reservierungen WHERE ReservierungsID = $resID";
$TITEL = "SELECT Titel FROM persons WHERE PersonID AND IS NOT NULL = $PID";
return $TITEL;
}
//##############################################
function giveHnummer($resID)
{
$PID = "SELECT PersonID FROM reservierungen WHERE ReservierungsID = $resID";
$TITEL = "SELECT Titel FROM persons WHERE PersonID AND IS NOT NULL = $PID";
return $TITEL;
}
//##############################################
function givePLZ($resID)
{
$PID = "SELECT PersonID FROM reservierungen WHERE ReservierungsID = $resID";
$TITEL = "SELECT Titel FROM persons WHERE PersonID AND IS NOT NULL = $PID";
return $TITEL;
}
//##############################################
function giveOrt($resID)
{
$PID = "SELECT PersonID FROM reservierungen WHERE ReservierungsID = $resID";
$TITEL = "SELECT Titel FROM persons WHERE PersonID AND IS NOT NULL = $PID";
return $TITEL;
}

Любая идея, почему это не работает?

заранее спасибо

docunhown