How to pass options to json_encode function in Symfony Serializer component - php

Question

here is the situation: I work on a rest api, based on symfony3, it uses FOSRestBundle and symfony serializer component, so methods return array and FOSRest handles encoding and response. The problem is serializer use json_encode with default settings and api return data like '\u00c9S' for some symbols. So I need to pass 'JSON_UNESCAPED_UNICODE' to json_encode() somehow. Is there any proper way to reach this goal?
Example of a method:
namespace AppBundle\Controller;
use Symfony\Bundle\FrameworkBundle\Controller\Controller;
use Sensio\Bundle\FrameworkExtraBundle\Configuration\Route;
use FOS\RestBundle\Controller\Annotations as Rest;
use Nelmio\ApiDocBundle\Annotation\ApiDoc;
use Symfony\Component\HttpFoundation\Request;
/**
* Class ExampleController
* #package AppBundle\Controller
*/
class ExampleController extends Controller
{
/**
* #Rest\Get("/get-some-data")
* #param Request $request
* #return array
*/
public function getSomeDataAction(Request $request)
{
$someData = [
'prop1' => 'Value',
'prop2' => 'Value',
'prop3' => 'Value',
'prop4' => 'Value',
];
return $someData;
}
So when I do request to '/get-some-data', it returns me:
{"prop1":"Value with \/","prop2":"Value with \u00c9"}
, but I need it to return:
{"prop1":"Value with /","prop2":"Value with É"}

Answer 1

I use Symfony 3 and the "Doctrine JSON ODM Bundle" to store my data as JSON document. I had the same problem. All the data that contained unicode characters where automatically escaped which was not what I wanted.
After some experiments I finally managed to pass JSON_UNESCAPED_UNICODE option to json_encode().
Below is my solution:
# config/services.yml
serializer.encode.json.unescaped:
class: Symfony\Component\Serializer\Encoder\JsonEncode
arguments:
- !php/const JSON_UNESCAPED_UNICODE
serializer.encoder.json:
class: Symfony\Component\Serializer\Encoder\JsonEncoder
arguments:
- '#serializer.encode.json.unescaped'

Answer 2

Solution for Symfony 5.
Passing an integer as first parameter of the "Symfony\Component\Serializer\Encoder\JsonEncode::__construct()" method is deprecated since Symfony 4.2, use the "json_encode_options" key of the context instead.
Add to config/services.yaml:
serializer.encode.json.unescaped:
class: Symfony\Component\Serializer\Encoder\JsonEncode
arguments:
- { "json_encode_options": !php/const JSON_UNESCAPED_UNICODE }
serializer.encoder.json:
class: Symfony\Component\Serializer\Encoder\JsonEncoder
arguments:
- '#serializer.encode.json.unescaped'

Answer 3

When set json type in doctrine for column under the hood is using PHP's json_encode() and json_decode() functions for storing and retrieving data from databases.
json_encode() function that convert PHP's array to JSON string has second parameter for assign set of options for set up some convertation rules. One of them is JSON_UNESCAPED_UNICODE that disable encoding multibyte symbols as \uXXXX
We can enable this option for json_encode() by default for doctrine. To levarage this you must do 2 steps.
1. Create class App\Doctrine\Types\JsonType
// App\Doctrine\Types\JsonType.php
<?php
namespace App\Doctrine\Types;
use Doctrine\DBAL\Platforms\AbstractPlatform;
class JsonType extends \Doctrine\DBAL\Types\JsonType
{
public function convertToDatabaseValue($value, AbstractPlatform $platform)
{
if ($value === null) {
return null;
}
$encoded = json_encode($value, JSON_UNESCAPED_UNICODE);
if (json_last_error() !== JSON_ERROR_NONE) {
throw ConversionException::conversionFailedSerialization($value, 'json', json_last_error_msg());
}
return $encoded;
}
}
2. Override configuration for DBAL JSON type corresponding to our class
# config/packages/doctrine.yaml
doctrine:
dbal:
types:
json: App\Doctrine\Types\JsonType

Answer 4

You can use an encoder, as found here and in the documentation
<?php
$encoder = new JsonEncoder(new JsonEncode(JSON_UNESCAPED_UNICODE), new JsonDecode(false));
$normalizer = new ObjectNormalizer();
$serializer = new Serializer(array($normalizer), array($encoder));
EDIT:
In this example, I use a Response object. Note that an Action Controller must return a Response object.
namespace AppBundle\Controller;
use Symfony\Bundle\FrameworkBundle\Controller\Controller;
use Sensio\Bundle\FrameworkExtraBundle\Configuration\Route;
use FOS\RestBundle\Controller\Annotations as Rest;
use Nelmio\ApiDocBundle\Annotation\ApiDoc;
use Symfony\Component\HttpFoundation\Request;
/**
* Class ExampleController
* #package AppBundle\Controller
*/
class ExampleController extends Controller
{
/**
* #Rest\Get("/get-some-data")
* #param Request $request
* #return array
*/
public function getSomeDataAction(Request $request)
{
$someData = [
'prop1' => 'Value',
'prop2' => 'Value',
'prop3' => 'Value',
'prop4' => 'Value',
];
$response = new Response($someData);
$response->headers->set('Content-Type', 'application/json');
return $response;
}

Answer 5

$this->serializer->serialize(
$data,
JsonEncoder::FORMAT,
['json_encode_options' => JSON_UNESCAPED_UNICODE]
);

Answer 6

In newer versions of symfony you have to use this format:
$encoders = [new JsonEncoder(new JsonEncode(["json_encode_options" =>
JSON_UNESCAPED_SLASHES]), new JsonDecode([]))];