更新时间:2023-11-22 21:27:34
ObjectNormalizer
需要更多配置.您至少需要提供PropertyTypeExtractorInterface
类型的第四个参数.
The ObjectNormalizer
needs more configuration. You will at least need to supply the fourth parameter of type PropertyTypeExtractorInterface
.
这是一个(相当hacky)的例子:
Here's a (rather hacky) example:
<?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,在示例json中,第一个条目具有一个数组作为make
的值.我认为这是一个错字,如果是故意的,请发表评论.
Note that in your example json, the first entry has an array as value for make
. I took this to be a typo, if it's deliberate, please leave a comment.
To make this more automatic you might want to experiment with the PhpDocExtractor
.