CODEWITHSUNDEEP
symfony-2.3jmsserializerbundle
Alex Bryanskiy
Alex Bryanskiy

Reputation: 445

How to expose property by JMS/Serializer for only one method?

I'm using JMS serializer in my Symfony projuect and i have a question about it. I want ot expose property from entity for only one specific method (one route), in other cases i dont want this property to be exposed. I would be appreciate for any advices)

Upvotes: 0

Views: 2251

Answers (1)

dlondero
dlondero

Reputation: 2595

You can probably achieve this using the @Groups annotation on your properties and then tell the serializer which groups to serialize in your controller.

use JMS\Serializer\Annotation\Groups;

class BlogPost
{
    /** @Groups({"list", "details"}) */
    private $id;

    /** @Groups({"list", "details"}) */
    private $title;

    /** @Groups({"list"}) */
    private $nbComments;

    /** @Groups({"details"}) */
    private $comments;

    private $createdAt;
}

And then:

use JMS\Serializer\SerializationContext;

$serializer->serialize(new BlogPost(), 'json', SerializationContext::create()->setGroups(array('list')));

//will output $id, $title and $nbComments.

$serializer->serialize(new BlogPost(), 'json', SerializationContext::create()->setGroups(array('Default', 'list')));

//will output $id, $title, $nbComments and $createdAt.

More info here.

Upvotes: 2

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