php - POST データをテーブルに挿入し、POST 配列を別のテーブルに挿入する

Question

フォームから 2 つの別々のテーブルにデータを送信しようとしています。

エラーは次のとおりです。テーブル 1 には正常に挿入されますが、テーブル 2 の配列データは " Array " としてデータベースに入ります。

テーブル1に入る私のフィールドは次のとおりです。

  $start = $_POST['start'];
  $end = $_POST['end'];
  $customer = $_POST['customer'];
  $manufacturer = $_POST['manufacturer'];
  $rep = $_POST['rep'];
  $notes = $_POST['notes'];

テーブル2に入る私の配列フィールド：

  item[]
  description[]
  pack[]

どんな助けでも大歓迎です。以下は、これまでに開発したコードです。

 if ($start == '' || $end == '')
                    {
                            $error = '<div class="alert alert-error">
                <a class="close" data-dismiss="alert">×</a>
                <strong>Error!</strong> Please fill in all required fields!
            </div>';
                    }
                    else
                    {
                        $sql = "SELECT COALESCE(MAX(GroupID), 0) + 1 AS newGroupID FROM table1";
try 
    { 
    $stmt = $db->prepare($sql); 
    $stmt->execute(); 
} 
catch(PDOException $ex) 
{ 
    die("Failed to run query: " . $ex->getMessage()); 
} 

$rows = $stmt->fetchAll(); 

foreach($rows as $row) {
$groupID = $row['newGroupID'];
}

$mysqli = new mysqli("localhost", "user", "pw", "mydb");
    if (mysqli_connect_errno()) {
        die(mysqli_connect_error());
    }


  $start = $_POST['start'];
  $end = $_POST['end'];
  $customer = $_POST['customer'];
  $manufacturer = $_POST['manufacturer'];
  $rep = $_POST['rep'];
  $notes = $_POST['notes'];

    if ($stmt = $mysqli->prepare("INSERT table1 (GroupID, start, end, customer, manufacturer, rep, notes) VALUES (?, ?, ?, ?, ?, ?, ?)"))
                            {
                                    $stmt->bind_param("issssss", $groupID, $start, $end, $customer, $manufacturer, $rep, $notes);
                                    $stmt->execute();
                                    $stmt->close();
                            }

                            else
                            {
                                    echo "ERROR: Could not prepare SQL statement 1.";
                            }


                            $mysqli->error;
                            $mysqli->close();
                            $success = "<div class='alert alert-success'>New agreement added.</div>";



                            $mysqli = new mysqli("localhost", "user", "pw", "mydb");
                            if (mysqli_connect_errno()) {
                                die(mysqli_connect_error());
                            }

                            if ($stmt = $mysqli->prepare("INSERT table2 (GroupID, item_number, item_description, pack) VALUES (?, ?, ?, ?)"))
                            {
                                    foreach($_POST['item'] as $i => $item) {
                                        $item = $_POST['item'];
                                        $description = $_POST['description'];
                                        $pack = $_POST['pack'];
                                    }

                                    $stmt->bind_param("isss", $GroupID, $item, $description, $pack);
                                    $stmt->execute();
                                    $stmt->close();
                            }

                            else
                            {
                                    echo "ERROR: Could not prepare SQL statement 2.";
                            }


                            $mysqli->error;
                            $mysqli->close();
                            $success = "<div class='alert alert-success'>New agreement items added!</div>";


            }
            }