1

我在 Payum 中实现了一个新的支付网关，我正在尝试更改通知操作中的响应，默认情况下 payum 发送 204 并且支付网关需要接收 200。

如何更改响应？

namespace xxxx\Bundle\xxxxxBundle\Pago\RedsysGateway\Action;

class StoreNotificationAction extends PaymentAwareAction
{


/**
 * {@inheritDoc}
 */
public function execute($request)
{

    /** @var $request SecuredNotifyRequest */
    if(!$this->supports($request)) {
        throw RequestNotSupportedException::createActionNotSupported($this, $request);
    }

    /** @var NotifyRequest $request */
    $notification = new NotificationDetails;
    $notification->setPaymentName($request->getToken()->getPaymentName());

    //save notification


}

/**
 * {@inheritDoc}
 */
public function supports($request)
{
    return
        $request instanceof SecuredNotifyRequest &&
        $request->getModel() instanceof Pago
        ;
}
}

这是 payum NotifyController：

namespace Payum\Bundle\PayumBundle\Controller;

use Payum\Core\Request\NotifyRequest;
use Payum\Core\Request\SecuredNotifyRequest;
use Symfony\Component\HttpFoundation\Request;
use Symfony\Component\HttpFoundation\Response;

class NotifyController extends PayumController
{
public function doUnsafeAction(Request $request)
{
    $payment = $this->getPayum()->getPayment($request->get('payment_name'));

    $payment->execute(new NotifyRequest(array_replace(
        $request->query->all(),
        $request->request->all()
    )));

    return new Response('', 204);
}

public function doAction(Request $request)
{
    $token = $this->getHttpRequestVerifier()->verify($request);

    $payment = $this->getPayum()->getPayment($token->getPaymentName());


    $payment->execute(new SecuredNotifyRequest(
        array_replace($request->query->all(), $request->request->all()),
        $token
    ));

    return new Response('', 204);
}
}
4

1 回答 1

1

您可以在NotifyAction.

class NotifyAction extends AbstractPaymentStateAwareAction
{
    public function execute($request)
    {
        // ...

        throw new ResponseInteractiveRequest(new Response('OK', 200));
    }

    public function supports($request)
    {
        return $request instanceof NotifyRequest;
    }
}

阅读有关交互式请求的https://github.com/Payum/Payum/blob/master/src/Payum/Core/Resources/docs/the-architecture.md

于 2014-06-05T06:08:09.817 回答