如何在symfony2中将cookie附加到JSON响应? [英] How to attach cookies to JSON Response in symfony2?

问题描述

我有JSON端点，用于将产品添加到购物车.它检查购物车是否已经存在.如果没有，它将创建一个购物车并将购物车ID存储在cookie中.因此，我如何将cookie附加到symfony2的JsonResponse上?

I have JSON endpoint which is used to add product to cart. It checks whether the cart already exists or no. If not then it creates a cart and the cart Id is stored in cookie. So I how do I attach cookie to the symfony2's JsonResponse ?

在非ajax版本中，如果我要从动作中渲染模板，则可以使用:

In a non ajax version if I am rendering a template from my action I can use:

$response = new Response();
$response->headers->setCookie(new Cookie(‘cookie_name’, ‘cookie_value’));

$this->render('<template_path>', '<array_options>', $response);

请帮助我了解如何针对JsonResponse进行操作.

Please help me on how to do it for a JsonResponse.

推荐答案

我认为，您会在以下链接下找到答案:

In my opinion, you will find the answer under the following links:

如何在symfony2控制器中发送json响应

http://api.symfony.com/2.2/Symfony/Component/HttpFoundation/JsonResponse.html

http://symfony.com/doc/current/components/http_foundation/introduction.html

查看链接和学习的最佳方法，但是如果找不到答案，也许会很好:

The best way, to see links and learn, but if you will not find the answer, maybe this will be good:

use Symfony\Component\HttpFoundation\Response;

$response = new Response();
$response->headers->set('Content-Type', 'application/json');
$response->headers->setCookie(new Cookie(‘cookie_name’, ‘cookie_value’));

return $response;

还有一个有用的JsonResponse类，它可以使这变得更加容易:

There is also a helpful JsonResponse class, which can make this even easier:

use Symfony\Component\HttpFoundation\JsonResponse;

$response = new JsonResponse();
$response->headers->setCookie(new Cookie(‘cookie_name’, ‘cookie_value’));

return $response;

我希望这会有所帮助:)

I hope, this will be helpfull :)

这篇关于如何在symfony2中将cookie附加到JSON响应?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！