我想使用Symfony XML序列化程序来转换类实例(而不是数组(。因此,例如,我想创建一个属性为myAtt="foo"的XML,
<?xml version="1.0"?>
<REQ>
<TravelAgencySender myAtt="foo">
<CityName>town</CityName>
<AgencyID>agency</AgencyID>
</TravelAgencySender>
</REQ>
所以我创建了一个这样的类
class TravelAgencySender
{
/**
* @var string
*/
private $CityName;
/**
* @var string
*/
public $AgencyID;
.....
}
和以下初始化
use SymfonyComponentSerializerEncoderJsonEncoder;
use SymfonyComponentSerializerNameConverterMetadataAwareNameConverter;
use SymfonyComponentSerializerNormalizerObjectNormalizer;
use SymfonyComponentSerializerSerializer;
use SymfonyComponentSerializerEncoderXmlEncoder;
use SymfonyComponentSerializerMappingFactoryClassMetadataFactory;
use SymfonyComponentSerializerMappingLoaderAnnotationLoader;
use DoctrineCommonAnnotationsAnnotationReader;
$classMetadataFactory = new ClassMetadataFactory(new AnnotationLoader(new AnnotationReader()));
$metadataAwareNameConverter = new MetadataAwareNameConverter($classMetadataFactory);
$serializer = new Serializer(
[new ObjectNormalizer($classMetadataFactory, $metadataAwareNameConverter)],
['json' => new JsonEncoder(), 'xml' => new XmlEncoder()]
);
有谁知道如何添加 myAtt 属性?
谢谢
这将产生以下 XML
<?xml version="1.0"?>
<REQ>
<TravelAgencySender>
<CityName>town</CityName>
<AgencyID>agency</AgencyID>
</TravelAgencySender>
</REQ>
以 @ 开头的数组键被视为 XML 属性:
['foo' => ['@bar' => 'value']];
编码如下:
<?xml version="1.0"?>
<response>
<foo bar="value"/>
</response>