问题描述
我正在使用 2.8 版的 Symfony 2 项目,并且我正在使用内置组件 Serializer ->http://symfony.com/doc/current/components/serializer.html
I'm working on a Symfony 2 project with version 2.8 and I'm using the build-in component Serializer -> 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,我想对对象中的内容进行非规范化.这是我的结构(汽车应用程序上下文中的模型/制造).
I have a JSON structure provided by a web service. After deserialization, I want to denormalize my content in objects. Here is my structure (model/make in a car application context).
[{
"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.
My idea is to populate a VehicleModel
object which contains a reference to a VehicleMake
object.
class VehicleModel {
public $id;
public $code;
public $model;
public $make; // VehicleMake
}
这是我所做的:
// Retrieve data in JSON
$data = ...
$serializer = new Serializer([new ObjectNormalizer(), new ArrayDenormalizer()], [new JsonEncoder()]);
$models = $serializer->deserialize($data, 'NamespaceVehicleModel[]', '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
代替.
In result, my object VehicleModel
is correctly populated but $make
is logically a key/value array. Here I want a VehicleMake
instead.
有没有办法做到这一点?
Is there a way to do that?
推荐答案
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(相当笨拙的)例子:
Here's a (rather hacky) example:
<?php
use SymfonyComponentPropertyInfoPropertyTypeExtractorInterface;
use SymfonyComponentPropertyInfoType;
use SymfonyComponentSerializerEncoderJsonEncoder;
use SymfonyComponentSerializerNormalizerArrayDenormalizer;
use SymfonyComponentSerializerNormalizerObjectNormalizer;
use SymfonyComponentSerializerSerializer;
$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);
<小时>
请注意,在您的示例 json 中,第一个条目有一个数组作为 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.
为了使更加自动化,您可能需要尝试使用 PhpDocExtractor
.
To make this more automatic you might want to experiment with the PhpDocExtractor
.
这篇关于使用 Symfony 2 序列化程序对对象中的嵌套结构进行非规范化的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,WP2