14

我有Product实体与多对一Category实体。我需要Product在会话中存储。首先，我尝试\Serializable在 Product 上实现接口。我应该如何序列化我的相关Category实体？我还应该实现\Serializable接口吗？

我读到，学说中的序列化是非常痛苦的操作，我想到了这一点：

我们可以从实体中获取原始值吗？正是存储在数据库中的数据。如果我们能得到这个值，我们就可以将它存储在任何地方并重新创建对象！

我阅读了学说 2Doctrine\ORM\Internal\Hydration\ObjectHydrator:hydrateRowData的代码并找到了方法，但它受到了保护。是否有任何公共 api 可以做到这一点？

更新：

我只是从 BasicEntityPersister 复制粘贴并集成了一些代码，它似乎可以工作。

    $product = $productsRepository->find($id);

    if (!$product) {
        throw $this->createNotFoundException('No product found for id ' . $id);
    }

    $uow = $em->getUnitOfWork();
    $entityPersister = $uow->getEntityPersister(get_class($product));
    $classMetadata = $entityPersister->getClassMetadata();

    $originalData = $uow->getOriginalEntityData($product);

    $result = array();
    foreach ($originalData as $field => $value) {
        if (isset($classMetadata->associationMappings[$field])) {
            $assoc = $classMetadata->associationMappings[$field];

            // Only owning side of x-1 associations can have a FK column.
            if ( ! $assoc['isOwningSide'] || ! ($assoc['type'] & \Doctrine\ORM\Mapping\ClassMetadata::TO_ONE)) {
                continue;
            }

            if ($value !== null) {
                $newValId = $uow->getEntityIdentifier($value);
            }

            $targetClass = $em->getClassMetadata($assoc['targetEntity']);
            $owningTable = $entityPersister->getOwningTable($field);

            foreach ($assoc['joinColumns'] as $joinColumn) {
                $sourceColumn = $joinColumn['name'];
                $targetColumn = $joinColumn['referencedColumnName'];

                if ($value === null) {
                    $result[$owningTable][$sourceColumn] = null;
                } else if ($targetClass->containsForeignIdentifier) {
                    $result[$owningTable][$sourceColumn] = $newValId[$targetClass->getFieldForColumn($targetColumn)];
                } else {
                    $result[$owningTable][$sourceColumn] = $newValId[$targetClass->fieldNames[$targetColumn]];
                }
            }
        } elseif (isset($classMetadata->columnNames[$field])) {
            $columnName = $classMetadata->columnNames[$field];
            $result[$entityPersister->getOwningTable($field)][$columnName] = $value;
        }
    }

    print_r($result);

在$result我们有原始值。现在我们需要方法如何通过这个数组创建对象

4

2 回答 2

7 
<?php

namespace Acme\ServiceBundle\Services;

use Doctrine\ORM\EntityManager;

class EntitySerializer
{
    protected $em;

    public function __construct(EntityManager $em)
    {
        $this->em = $em;
    }

    public function serialize($entity)
    {
        $className = get_class($entity);

        $uow = $this->em->getUnitOfWork();
        $entityPersister = $uow->getEntityPersister($className);
        $classMetadata = $entityPersister->getClassMetadata();

        $result = array();
        foreach ($uow->getOriginalEntityData($entity) as $field => $value) {
            if (isset($classMetadata->associationMappings[$field])) {
                $assoc = $classMetadata->associationMappings[$field];

                // Only owning side of x-1 associations can have a FK column.
                if ( ! $assoc['isOwningSide'] || ! ($assoc['type'] & \Doctrine\ORM\Mapping\ClassMetadata::TO_ONE)) {
                    continue;
                }

                if ($value !== null) {
                    $newValId = $uow->getEntityIdentifier($value);
                }

                $targetClass = $this->em->getClassMetadata($assoc['targetEntity']);
                $owningTable = $entityPersister->getOwningTable($field);

                foreach ($assoc['joinColumns'] as $joinColumn) {
                    $sourceColumn = $joinColumn['name'];
                    $targetColumn = $joinColumn['referencedColumnName'];

                    if ($value === null) {
                        $result[$sourceColumn] = null;
                    } else if ($targetClass->containsForeignIdentifier) {
                        $result[$sourceColumn] = $newValId[$targetClass->getFieldForColumn($targetColumn)];
                    } else {
                        $result[$sourceColumn] = $newValId[$targetClass->fieldNames[$targetColumn]];
                    }
                }
            } elseif (isset($classMetadata->columnNames[$field])) {
                $columnName = $classMetadata->columnNames[$field];
                $result[$columnName] = $value;
            }
        }

        return array($className, $result);
    }

    public function deserialize(Array $data)
    {
        list($class, $result) = $data;

        $uow = $this->em->getUnitOfWork();
        return $uow->createEntity($class, $result);
    }
}
于 2012-11-23T02:36:15.407 回答
6

我遇到了同样的问题，我也想关联数组中的数据。所以我想出了以下内容：

$serializer = new Serializer($this->em); // Pass the EntityManager object
$array = $serializer->serialize($message); // Returns the array (with associations!)

资源：

<?php

/**
 * Class Serializer
 *
 * @author Steffen Brem
 */
class Serializer
{
    /**
     * @var Doctrine\ORM\EntityManager
     */
    private $_em;

    /**
     * Constructor
     *
     * @param \Doctrine\ORM\EntityManager $em
     */
    public function __construct(\Doctrine\ORM\EntityManager $em)
    {
        $this->_em = $em;
    }

    /**
     * Serialize entity to array
     *
     * @param $entityObject
     * @return array
     */
    public function serialize($entityObject)
    {
        $data = array();

        $className = get_class($entityObject);
        $metaData = $this->_em->getClassMetadata($className);

        foreach ($metaData->fieldMappings as $field => $mapping)
        {
            $method = "get" . ucfirst($field);
            $data[$field] = call_user_func(array($entityObject, $method));
        }

        foreach ($metaData->associationMappings as $field => $mapping)
        {
            // Sort of entity object
            $object = $metaData->reflFields[$field]->getValue($entityObject);

            $data[$field] = $this->serialize($object);
        }

        return $data;
    }
}
于 2013-07-06T13:49:22.940 回答