Question

我知道它反对框架使实体容器知道，但这是一个特例，我有一个信用卡实体，我想这样做：

/**
 * @return mixed
 */
public function getNumber()
{
    $number = $this->number;
    $crypt = base64_decode($number);
    $number = mcrypt_decrypt(MCRYPT_RIJNDAEL_256, $key, $crypt, MCRYPT_MODE_ECB);
    return trim($number);
}

/**
 * @param $number
 * @return $this
 */
public function setNumber($number)
{
    $crypt = mcrypt_encrypt(MCRYPT_RIJNDAEL_256, $key, $number, MCRYPT_MODE_ECB);
    $number = trim(base64_encode($crypt));
    $this->number = $number;
    return $this;
}

我希望$ key是来自parameters.yml的秘密，因为我不想将它保存在代码中。

当我使用FormType时，我无法将其作为参数传递，因为表单类型在绑定请求时不会传递它。

    $credit_card = new CreditCard();
    $credit_card->setCustomer($customer);
    $payment_form = $this->createPaymentForm($credit_card);

    $payment_form->handleRequest($request);

Answer 1

从您的实体中删除数据修改，并将其添加到您的FormType。

public function getNumber()
{
    return $this->number;
}

public function setNumber($number)
{
    $this->number = $number;

    return $this;
}

让你的FormType将参数作为参数。

class CreditCardType extends AbstractType {

    private $key;

    public function __construct($key) {
        $this->key = $key;
    }

    // ...

    // Listen on the PRE_BIND event to update your value before binding
    $builder->addEventListener(FormEvents::PRE_BIND, function (FormEvent $event) 
    {
        $data = $event->getData();
        $number = $data['number'];
        $crypt = $this->encryptNumber($number);
        $data['number'] = $crypt;

        $event->setData($data);
    });

    public function encryptNumber($key, $number)
    {
        $crypt = mcrypt_encrypt(MCRYPT_RIJNDAEL_256, $key, $number, MCRYPT_MODE_ECB);
        $number = trim(base64_encode($crypt));

        return $number;    
    }
}

现在，在您的createPaymentForm中，将参数添加到FormType实例中，如下所示：

$param = $this->container->getParameter('yourkeyparam'); 
$form = $this->createForm(new CreditCardType($param), $credit_card, array());

希望这就是你所需要的。

Symfony2在实体中获取parameters.yml，特殊情况

1 个答案: