Question

我正在使用版本2.8的Symfony 2项目，而我正在使用内置组件Serializer - ＆gt; http://symfony.com/doc/current/components/serializer.html

我有一个由Web服务提供的JSON结构。反序列化后，我想在对象中反规范化我的内容。这是我的结构（汽车应用程序上下文中的模型/品牌）。

[{
"0": {
    "id": 0,
    "code": 1,
    "model": "modelA",
    "make": {
        "id": 0,
        "code": 1,
        "name": "makeA"
    }
  }
} , {
 "1": {
    "id": 1,
    "code": 2,
    "model": "modelB",
    "make": {
        "id": 0,
        "code": 1,
        "name": "makeA"
    }
  }
}]

我的想法是填充一个VehicleModel对象，其中包含对VehicleMake对象的引用。

class VehicleModel {
    public $id;
    public $code;
    public $model;
    public $make; // VehicleMake
}

以下是我的工作：

// Retrieve data in JSON
$data = ...
$serializer = new Serializer([new ObjectNormalizer(), new ArrayDenormalizer()], [new JsonEncoder()]);
$models = $serializer->deserialize($data, '\Namespace\VehicleModel[]', 'json');

结果，我的对象VehicleModel被正确填充，但$make在逻辑上是一个键/值数组。在这里，我想要一个VehicleMake。

有办法吗？

THX

Answer 1

ObjectNormalizer需要更多配置。您至少需要提供PropertyTypeExtractorInterface类型的第四个参数。

这是一个（相当hacky）的例子：

<?php
use Symfony\Component\PropertyInfo\PropertyTypeExtractorInterface;
use Symfony\Component\PropertyInfo\Type;
use Symfony\Component\Serializer\Encoder\JsonEncoder;
use Symfony\Component\Serializer\Normalizer\ArrayDenormalizer;
use Symfony\Component\Serializer\Normalizer\ObjectNormalizer;
use Symfony\Component\Serializer\Serializer;

$a = new VehicleModel();
$a->id = 0;
$a->code = 1;
$a->model = 'modalA';
$a->make = new VehicleMake();
$a->make->id = 0;
$a->make->code = 1;
$a->make->name = 'makeA';

$b = new VehicleModel();
$b->id = 1;
$b->code = 2;
$b->model = 'modelB';
$b->make = new VehicleMake();
$b->make->id = 0;
$b->make->code = 1;
$b->make->name = 'makeA';

$data = [$a, $b];

$serializer = new Serializer(
    [new ObjectNormalizer(null, null, null, new class implements PropertyTypeExtractorInterface {
        /**
         * {@inheritdoc}
         */
        public function getTypes($class, $property, array $context = array())
        {
            if (!is_a($class, VehicleModel::class, true)) {
                return null;
            }

            if ('make' !== $property) {
                return null;
            }

            return [
                new Type(Type::BUILTIN_TYPE_OBJECT, true, VehicleMake::class)
            ];
        }
    }), new ArrayDenormalizer()],
    [new JsonEncoder()]
);

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

$models = $serializer->deserialize($json, VehicleModel::class . '[]', 'json');
print_r($models);

请注意，在您的示例json中，第一个条目的数组为make的值。我认为这是一个错字，如果它是故意的，请发表评论。

要使此更自动，您可能需要尝试使用PhpDocExtractor。

Answer 2

如果您需要更多的非规范化灵活性，那么创建自己的非规范化器会很好。

$serializer = new Serializer(
  [
    new ArrayNormalizer(), 
    new VehicleDenormalizer(), 
    new VehicleMakeDenormalizer()
  ], [
    new JsonEncoder()
  ]
);
$models = $serializer->deserialize(
  $data, 
  '\Namespace\VehicleModel[]', 
  'json'
);

这里是这种非规范化程序的粗略代码

class VehicleDenormalizer implements DenormalizerInterface, DenormalizerAwareInterface
    {
      public function denormalize($data, $class, $format, $context) 
      {
        $vehicle = new VehicleModel();
        ...
        $vehicleMake = $this->denormalizer->denormalize(
          $data->make,
          VehicleMake::class,
          $format,
          $context
        );
        $vehicle->setMake($vehicleMake);
        ...
      }
    }

我只怀疑我们是否应该依赖$this->denormalizer->denormalize（因为我们使用Symfony\Component\Serializer\Serializer而正常工作）或者我们必须明确地将VehicleMakeDenormalizer注入VehicleDenormalizer

$vehicleDenormalizer = new VehicleDenormalizer();
$vehicleDenormalizer->setVehicleMakeDenormalizer(new VehicleMakeDenormalizer());

Answer 3

如果您的Vehicle类具有一些类型提示，最简单的方法是使用ReflectionExtractor。

class VehicleModel {
    public $id;
    public $code;
    public $model;
    /** @var VehicleMake */
    public $make;
}

初始化Symfony\Component\PropertyInfo\Extractor\ReflectionExtractor时，可以将ObjectNormalizer作为参数传递给Serializer

$serializer = new Serializer([new ObjectNormalizer(null, null, null, new ReflectionExtractor()), new ArrayDenormalizer()], [new JsonEncoder()]);
$models = $serializer->deserialize($data, '\Namespace\VehicleModel[]', 'json');

Answer 4

在 Symfony4+ 中，您可以注入序列化程序，它会根据您的 phpdoc（例如 @var）或类型提示为您完成工作。 Phpdoc 似乎更安全，因为它管理对象集合。

示例：

App\Model\Skill.php

<?php

namespace App\Model;

class Skill
{
    public $name = 'Taxi Driver';

    /** @var Category */
    public $category;

    /** @var Person[] */
    public $people = [];
}

App\Model\Category.php

<?php

namespace App\Model;

class Category
{
    public $label = 'Transports';
}

App\Model\Person.php

<?php

namespace App\Model;

class Person
{
    public $firstname;
}

App\Command\TestCommand.php

<?php

namespace App\Command;

use App\Model\Category;
use App\Model\Person;
use App\Model\Skill;
use Symfony\Component\Console\Command\Command;
use Symfony\Component\Console\Input\InputInterface;
use Symfony\Component\Console\Output\OutputInterface;
use Symfony\Component\Serializer\SerializerInterface;

class TestCommand extends Command
{
    /**
     * @var SerializerInterface
     */
    private $serializer;

    public function __construct(SerializerInterface $serializer)
    {
        parent::__construct();

        $this->serializer = $serializer;
    }

    protected function configure()
    {
        parent::configure();

        $this
            ->setName('test')
            ->setDescription('Does stuff');
    }

    protected function execute(InputInterface $input, OutputInterface $output)
    {
        $personA            = new Person();
        $personA->firstname = 'bruno';
        $personB            = new Person();
        $personB->firstname = 'alice';

        $badge           = new Skill();
        $badge->name     = 'foo';
        $badge->category = new Category();
        $badge->people   = [$personA, $personB];

        $output->writeln(
            $serialized = $this->serializer->serialize($badge, 'json')
        );

        $test = $this->serializer->deserialize($serialized, Skill::class, 'json');

        dump($test);

        return 0;
    }
}

将给出以下预期结果：

{"name":"foo","category":{"label":"Transports"},"people":[{"firstname":"bruno"},{"firstname":"alice"}]}

^ App\Model\BadgeFacade^ {#2531
  +name: "foo"
  +category: App\Model\CategoryFacade^ {#2540
    +label: "Transports"
  }
  +people: array:2 [
    0 => App\Model\PersonFacade^ {#2644
      +firstname: "bruno"
    }
    1 => App\Model\PersonFacade^ {#2623
      +firstname: "alice"
    }
  ]
}

使用symfony 2序列化器对对象中的嵌套结构进行非规范化

4 个答案: