Symfony Restful API - Expose virtual property isLiked by current logged in user - php

Question

There are two entities Restaurant and Users. Restaurant entity has many-to-many relation with user, field name favoriteBy.
<many-to-many field="favoriteBy" target-entity="UserBundle\Entity\Users" mapped-by="favoriteRestaurants"/>
I am using JMS Serializer along with FOSRestfulAPI. In restaurant listing API I have to expose one extra boolean field "isFavorited", which will be true if current logged in user has in array collection favoriteBy.
How I can find whether current user has favorited the restaurant or not within entity?
/**
* Get is favorited
* #JMS\VirtualProperty()
* #JMS\Groups({"listing", "details"})
*/
public function isFavorited()
{
// some logic in entity
return false;
}
One way I am thinking is to inject current user object to entity and user contains method to find out, but its look like not good approach.
Please suggest me some method, or guide me to right direction.

Answer 1

You could implments an EventSubscriberInterface as described here in the doc.
As Example:
use JMS\Serializer\EventDispatcher\EventSubscriberInterface;
use JMS\Serializer\EventDispatcher\ObjectEvent;
...
class RestaurantSerializerSubscriber implements EventSubscriberInterface
{
protected $tokenStorage;
public function __construct(TokenStorageInterface $tokenStorage)
{
$this->tokenStorage = $tokenStorage;
}
public static function getSubscribedEvents()
{
return [
[
'event' => 'serializer.post_serialize',
'class' => Restaurant::class,
'method' => 'onPostSerialize',
],
];
}
public function onPostSerialize(ObjectEvent $event)
{
$visitor = $event->getVisitor();
$restaurant = $event->getObject();
// your custom logic
$isFavourite = $this->getCurrentUser()->isFavourite($restaurant);
$visitor->addData('isFavorited', $isFavourite);
}
/**
* Return the logged user.
*
* #return User
*/
protected function getCurrentUser()
{
return $this->tokenStorage->getToken()->getUser();
}
And register, as YML example:
acme.restaurant_serializer_subscriber:
class: Acme\DemoBundle\Subscriber\RestaurantSerializerSubscriber
arguments: ["#security.token_storage"]
tags:
- { name: "jms_serializer.event_subscriber" }
Hope this help
PS: You could also intercept the serialization group selected, let me know if you neet that code.

Answer 2

Entity should know nothing about current logged in user so injecting user into entity is not a good idea.
Solution 1:
This can be done with custom serialization:
// serialize single entity or collection
$data = $this->serializer->serialize($restaurant);
// extra logic
$data['is_favourited'] = // logic to check if it's favourited by current user
// return serialized data
Solution 2
This can be also achieved by adding Doctrine2->postLoad listener or subscriber after loading Restaurant entity. You can add dependency for current authenticated token to such listener and set there Restaurant->is_favorited virtual property that will be next serialized with JMS.