prepared insert query kan niet debuggen

ini_set('display_errors', 1);
error_reporting(E_ALL); 
$newBlog = $DBD->conn()->prepare('INSERT INTO woningen (iso,active,aangepast,title,seo_title,description,lat,lng,region2,region2url,region3,region3url,address,amt_images,rooms,price_rent,price_buy,house_type,size,broker,build_year,status,buurt,bouwvorm) VALUES (?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?)' ) or die (mysqli_error($DBD->conn()));
										$newBlog->bind_param( "ssssssssssssssssssssssss",$iso,$active,$aangepast,$title, $seo_title,$desc,$lat,$lng,$rt['Region2'],$rt['Region2Url'],$rt['Region3'],$rt['Region3Url'],$adres,$amt_images,$rooms,$huurprijs,$vraagprijs,$house_type,$opp,$mak_id ,$bouwjaar,$status,$buurt,$bouwvorm );
									
									
									echo $iso,$active,$aangepast,$title, $seo_title,$desc,$lat,$lng,$rt['Region2'],$rt['Region2Url'],$rt['Region3'],$rt['Region3Url'],$adres,$amt_images,$rooms,$huurprijs,$vraagprijs,$house_type,$opp,$mak_id ,$bouwjaar,$status,$buurt,$bouwvorm ;
										if (!$newBlog->execute()) 
										{
													echo 'error executing statement: ' . $newBlog->error;
													$error = 1;
													die();
										} else {
										echo 'goed';
										echo  $newBlog->error;;
										exit;
										$last_id = mysqli_insert_id($DBD->conn());	
										}