Symfony Serializer of DateTime Response String

Question

When attempting to perform a doctrine query and serialze it to json (not using JSM, using the symfony serializer) I get the following in the json:

""due":{"timezone":{"name":"Europe/Berlin","transitions":[{"ts":-2147483648,"time":"1901-12-13T20:45:52+0000","offset":3208,"isdst":false,"abbr":"LMT"},{"ts":-2147483648,"time":"1901-12-13T20:45:52+0000","offset":3600,"isdst":false,"abbr":"CET"},{"ts":-1693706400,"time":"1916-04-30T22:00:00+0000","offset":7200,"isdst":true,"abbr":"CEST"},{"ts":-1680483600,"time":"1916-09-30T23:00:00+0000","offset":3600,"isdst":false,"abbr":"CET"},{"ts":-1663455600,"time":"1917-04-16T01:00:00+0000","offset":7200,"isdst":true,"abbr":"CEST"},{"ts":-1650150000,"time":"1917-09-17T01:00:00+0000","offset":3600,"isdst":false,"abbr":"CET"},{"ts":-1632006000,"time":"1918-04-15T01:00:00+0000","offset":7200,"isdst":true,"abbr":"CEST"},{"ts":-1618700400,"time":"1918-09-16T01:00:00+0000","offset":3600,"isdst":false,"abbr":"CET"},{"ts":-938905200,"time":"1940-04-01T01:00:00+0000","offset":7200,"isdst":true,"abbr":"CEST"},{"ts":-857257200,"time":"1942-11-02T01:00:00+0000","offset":3600,"isdst":false,"abbr":"CET"},{"ts":-844556400,"time":"1943-03-29T01:00:00+0000","offset":7200,"isdst":true,"abbr":"CEST"},{"ts":-828226800,"time":"1943-10-04T01:00:00+0000","offset":3600,"isdst":false,"abbr":"CET"},{"ts":-812502000,"time":"1944-04-03T01:00:00+0000","offset":7200,"isdst":true,"abbr":"CEST"},{"ts":-796777200,"time":"1944-10-02T01:00:00+0000","offset":3600,"

When storing the due date, the timezone is saved and there is an additional timezone field stored. Is there a way to pull the date in a specific format or specify the timezone to use when retrieving?

 public function blahAction(Request $request)
{
    $currentUser = $this->getUser();
    $em = $this->getDoctrine()->getManager();
    $blahs = $em->getRepository('AppBundle:blah')->findAllByStatus($currentUser,'TODO');
    $encoders = array(new XmlEncoder(), new JsonEncoder());
    $normalizer = array(new ObjectNormalizer());
    $serializer = new Serializer($normalizer, $encoders);
    $response = new Response($serializer->serialize($blahs, 'json'));
    $response->headers->set('Content-Type', 'application/json');
    return $response;
}

/** * @ORM\Column(type="datetimetz", nullable=true) */ private $due; — Lee
– Lee, Commented Apr 11, 2018 at 8:59
Thanks. You must add DateTimeNormalizer. Check my answer. I prefer using Serializer service. — SilvioQ
– SilvioQ, Commented Apr 11, 2018 at 16:52

SilvioQ · Accepted Answer · 2018-04-11 16:51:34Z

17

You have 2 ways to get RFC3339 Datetime format ...

Option 1:

Add DateTimeNormalizer as normalizer. An example is https://symfony.com/doc/current/components/serializer.html#recursive-denormalization-and-type-safety.

Change

$normalizer = array(new ObjectNormalizer());

by

$normalizer = array(new DateTimeNormalizer(), new ObjectNormalizer());

Option 2

More simple is using "serializer" container service ... your code will look like ...

public function blahAction(Request $request)
{
    $currentUser = $this->getUser();
    $em = $this->getDoctrine()->getManager();
    $blahs = $em->getRepository('AppBundle:blah')->findAllByStatus($currentUser,'TODO');
    $response = new Response($this->get('serializer')->serialize($blahs, 'json'));
    $response->headers->set('Content-Type', 'application/json');
    return $response;
}

.. or, if your prefer autowiring way (this code is unchecked)

public function blahAction(Request $request, \Symfony\Component\Serializer\SerializerInterface $serializer)
{
    $currentUser = $this->getUser();
    $em = $this->getDoctrine()->getManager();
    $blahs = $em->getRepository('AppBundle:blah')->findAllByStatus($currentUser,'TODO');
    $response = new Response($serializer->serialize($blahs, 'json'));
    $response->headers->set('Content-Type', 'application/json');
    return $response;
}

answered Apr 11, 2018 at 16:51

SilvioQ

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1 Comment

Frantisek Over a year ago

Just note.. when defining $normalizer array, DateTimeNormalizer() should be first. Order matters.

Igor Shumichenko · Accepted Answer · 2020-08-18 18:24:45Z

In Symfony 5.1 it's recomended to add a callback function to specify datetime field format.For example:

use Symfony\Component\Serializer\Serializer;
use Symfony\Component\Serializer\Encoder\JsonEncoder;
use Symfony\Component\Serializer\Normalizer\ObjectNormalizer;
use Symfony\Component\Serializer\Normalizer\AbstractNormalizer;


$dateCallback = function ($innerObject, $outerObject, string $attributeName, string $format = null, array $context = []) {
        return $innerObject instanceof \DateTime ? $innerObject->format(\DateTime::ISO8601) : '';
    };

    $defaultContext = [
        AbstractNormalizer::CALLBACKS => [
            'created_at' => $dateCallback,
            'updated_at' => $dateCallback
        ],
        AbstractNormalizer::CIRCULAR_REFERENCE_HANDLER =>
            function ($articles, $format, $context)  {
                return $articles->getId();
            }
    ];

    $encoders = [new JsonEncoder()];
    $normalizers = [
        new ObjectNormalizer(null, null, null,
            null, null, null,
            $defaultContext)
    ];

    $serializer = new Serializer(
        $normalizers, $encoders
    );

    $articles = $serializer->serialize($articles, 'json');

Where $articles = array of App\Entity\Article objects

You can specify datetime format you need in your callback function.

Ilya Levin · Accepted Answer · 2020-11-21 14:19:37Z

3

Another option is to pass arguments to the default DateTimeNormalizer in services.yaml:

  Symfony\Component\Serializer\Normalizer\DateTimeNormalizer:
    arguments:
      $defaultContext:
        datetime_format: 'Y-m-d\TH:i:s.v\Z'

answered Nov 21, 2020 at 14:19

Ilya Levin

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