参加以下 DTO 课程:
class UserDTO {
/**
* @param AddressDTO[] $addressBook
*/
public function __construct(
public string $name,
public int $age,
public ?AddressDTO $billingAddress,
public ?AddressDTO $shippingAddress,
public array $addressBook,
) {
}
}
class AddressDTO {
public function __construct(
public string $street,
public string $city,
) {
}
}
我想将它们序列化为 JSON 或反序列化为 JSON。
我正在使用以下序列化器配置:
$encoders = [new JsonEncoder()];
$extractor = new PropertyInfoExtractor([], [
new PhpDocExtractor(),
new ReflectionExtractor(),
]);
$normalizers = [
new ObjectNormalizer(null, null, null, $extractor),
new ArrayDenormalizer(),
];
$serializer = new Serializer($normalizers, $encoders);
但是当序列化/反序列化这个对象时:
$address = new AddressDTO('Rue Paradis', 'Marseille');
$user = new UserDTO('John', 25, $address, null, [$address]);
$jsonContent = $serializer->serialize($user, 'json');
dd($serializer->deserialize($jsonContent, UserDTO::class, 'json'));
我得到以下结果:
UserDTO^ {#54
+name: "John"
+age: 25
+billingAddress: AddressDTO^ {#48
+street: "Rue Paradis"
+city: "Marseille"
}
+shippingAddress: null
+addressBook: array:1 [
0 => array:2 [
"street" => "Rue Paradis"
"city" => "Marseille"
]
]
}
当我期望时:
UserDTO^ {#54
+name: "John"
+age: 25
+billingAddress: AddressDTO^ {#48
+street: "Rue Paradis"
+city: "Marseille"
}
+shippingAddress: null
+addressBook: array:1 [
0 => AddressDTO^ {#48
+street: "Rue Paradis"
+city: "Marseille"
}
]
}
如您所见,
$addressBook
被反序列化为array
数组,而不是AddressDTO
数组。 我希望 PhpDocExtractor
从构造函数中读取 @param AddressDTO[]
,但这不起作用。
仅当我将
$addressBook
设为用 @var
记录的公共财产时,它才有效。
有没有办法让它在构造函数上使用简单的
@param
?
(非)工作演示:https://phpsandbox.io/n/gentle-mountain-mmod-rnmqd
我读过并尝试过的:
所提出的解决方案似乎都不适合我。
显然问题在于
PhpDocExtractor
不从构造函数中提取属性。为此,您需要使用特定的提取器:
use Symfony\Component\PropertyInfo;
use Symfony\Component\Serializer;
$phpDocExtractor = new PropertyInfo\Extractor\PhpDocExtractor();
$typeExtractor = new PropertyInfo\PropertyInfoExtractor(
typeExtractors: [ new PropertyInfo\Extractor\ConstructorExtractor([$phpDocExtractor]), $phpDocExtractor,]
);
$serializer = new Serializer\Serializer(
normalizers: [
new Serializer\Normalizer\ObjectNormalizer(propertyTypeExtractor: $typeExtractor),
new Serializer\Normalizer\ArrayDenormalizer(),
],
encoders: ['json' => new Serializer\Encoder\JsonEncoder()]
);
这样你就会得到想要的结果。我花了一点时间才弄清楚。多个反规范化器/提取器链总是让我着迷。
或者,对于更复杂的操作系统特殊情况,您可以创建自己的自定义反规范化器:
use Symfony\Component\Serializer\Normalizer\DenormalizerInterface;
use Symfony\Component\Serializer\Normalizer\DenormalizerAwareInterface;
use Symfony\Component\Serializer\Normalizer\DenormalizerAwareTrait
class UserDenormalizer
implements DenormalizerInterface, DenormalizerAwareInterface
{
use DenormalizerAwareTrait;
public function denormalize($data, string $type, string $format = null, array $context = [])
{
$addressBook = array_map(fn($address) => $this->denormalizer->denormalize($address, AddressDTO::class), $data['addressBook']);
return new UserDTO(
name: $data['name'],
age: $data['age'],
billingAddress: $this->denormalizer->denormalize($data['billingAddress'], AddressDTO::class),
shippingAddress: $this->denormalizer->denormalize($data['shippingAddress'], AddressDTO::class),
addressBook: $addressBook
);
}
public function supportsDenormalization($data, string $type, string $format = null)
{
return $type === UserDTO::class;
}
}
设置将变成这样:
$extractor = new PropertyInfoExtractor([], [
new PhpDocExtractor(),
new ReflectionExtractor(),
]);
$userDenormalizer = new UserDenormalizer();
$normalizers = [
$userDenormalizer,
new ObjectNormalizer(null, null, null, $extractor),
new ArrayDenormalizer(),
];
$serializer = new Serializer($normalizers, [new JsonEncoder()]);
$userDenormalizer->setDenormalizer($serializer);
输出成为您所期望的:
^ UserDTO^ {#39
+name: "John"
+age: 25
+billingAddress: AddressDTO^ {#45
+street: "Rue Paradis"
+city: "Marseille"
}
+shippingAddress: null
+addressBook: array:2 [
0 => AddressDTO^ {#46
+street: "Rue Paradis"
+city: "Marseille"
}
]
}