Bonjour à tous.
Mon code Php est non maitrisé.
Ce code ne marche pas. Pourquoi ?
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if (isset($_POST['identifiant'])) {
    $order_number = $_POST['identifiant'];
    $employeur = $_POST['employeur'];
    $activite = $_POST['activite'];
    $salaire = $_POST['salaire'];
    $debutActivite = $_POST['debutActivite'];
    $finActivite = $_POST ['finActivite'];
    try {
$host = '127.0.0.1';
        $db   = 'philippe';
        $user = 'root';
        $pass = '';
        $port = "3306";
        $charset = 'utf8mb4';
 
        $options = [
            \PDO::ATTR_ERRMODE            => \PDO::ERRMODE_EXCEPTION,
            \PDO::ATTR_DEFAULT_FETCH_MODE => \PDO::FETCH_ASSOC,
            \PDO::ATTR_EMULATE_PREPARES   => false,
        ];
        $dsn = "mysql:host=$host;dbname=$db;charset=$charset;port=$port";
        try {
            $pdo = new \PDO($dsn, $user, $pass, $options);
        } catch (\PDOException $e) {
            throw new \PDOException($e->getMessage(), (int)$e->getCode());
        }
 
//       $pdo = new PDO('mysql:host=localhost;dbname=philippe','root','');
 
      $data = [
          'employeur' => $employeur,
          'activite' => $activite,
          'salaire' => $salaire,
          'debutActivite' => $debutActivite,
          'finActivite' => $finActivite,
          'order_number' => $order_number,
              ];
 
      $sql = "UPDATE orders SET employeur = :employeur,activite = :activite, salaire = :salaire, debutActivite = :debutActivite,
finActivite =: finActivite  where order_number =: order_number";
 
      $stmt= $pdo->prepare($sql);
      $stmt->execute($data);  
 
    } catch (Exception $e) { echo "Erreure";}
 
    echo "Success";
Je ne sais pas ou ça bug.
Merci.