php - OneToMany 或 OneToOne，我走的路对还是错？

Question

我有这个数据库模型:

然后我做了这个实体(我只留下关系部分，因为另一个与主题不相关):

Orders.php

class Orders {
    /**
     * @ORM\ManyToOne(targetEntity="Person", inversedBy="orders")
     * @ORM\JoinColumn(name="person_id", referencedColumnName="id")
     * */
    protected $person;

    public function setPerson(Person $person)
    {
        $this->person = $person;
        return $this;
    }

    public function getPerson()
    {
        return $this->person;
    }

}

Person.php

class Person {
    /**
     * @ORM\OneToMany(targetEntity="NaturalPerson", mappedBy="person")
     * */
    private $naturals;

    /**
     * @ORM\OneToMany(targetEntity="LegalPerson", mappedBy="person")
     * */
    private $legals;

    /**
     * @ORM\OneToMany(targetEntity="Orders", mappedBy="person")
     * */
    private $orders;

    public function __construct()
    {
        $this->naturals = new ArrayCollection();
        $this->legals = new ArrayCollection();
        $this->orders = new ArrayCollection();
    }

    public function getNaturals()
    {
        return $this->naturals;
    }

    public function getLegals()
    {
        return $this->legals;
    }

    public function getOrders()
    {
        return $this->orders;
    }

}

NaturalPerson.php

class NaturalPerson {

    /**
     * @ORM\Id
     * @ORM\ManyToOne(targetEntity="Person", inversedBy="naturals")
     * @ORM\JoinColumn(name="person_id", referencedColumnName="id")
     */
    protected $person;

    /**
     * @ORM\Column(name="identification_type", type="ci_type", nullable=false)
     * @DoctrineAssert\Enum(entity="Tanane\FrontendBundle\DBAL\Types\CIType")
     */
    protected $identification_type;

    /**
     * @ORM\Column(name="ci", type="integer", nullable=false)
     */
    protected $ci;

    public function setPerson(Person $person)
    {
        $this->person = $person;
        return $this;
    }

    public function getPerson()
    {
        return $this->person;
    }

    public function setIdentificationType($identification_type)
    {
        $this->identification_type = $identification_type;
        return $this;
    }

    public function getIdentificationType()
    {
        return $this->identification_type;
    }

    public function setCI($ci)
    {
        $this->ci = $ci;
        return $this;
    }

    public function getCI()
    {
        return $this->ci;
    }

}

我省略了 LegalPerson，因为它与 NaturalPerson 几乎相同，所以这就是问题所在。映射看起来不错，但我如何从 Orders 获取相关记录？

这背后的想法是对于每个 Orders 我需要知道哪个 Person 也属于(Orders)以及存储在 NaturalPerson< 的额外信息 或 LegalPerson 取决于 person.type。

查看这段代码:

public function getOrdersAction()
{
    $response = array();
    $em = $this->getDoctrine()->getManager();

    $entities = $em->getRepository("FrontendBundle:Orders")->findAll();

    if (!$entities)
    {
        $response['message'] = "No se encontraron resultados";
    }

    $orders = array();
    foreach ($entities as $entity)
    {

        $personType = $entity->getPerson()->getPersonType();

        $order = array();
        $order[] = $entity->getNickname();

        // Here I'm trying to access to `Naturals` methods from `Orders` 
        if ($personType == 1)
        {
            $order[] = $entity->getPerson()->getNaturals()[0]->getIdentificationType() . $entity->getPerson()->getNaturals()[0]->getCI();
        }
        elseif ($personType == 2)
        {
            $order[] = $entity->getPerson()->getLegals()[0]->getIdentificationType() . $entity->getPerson()->getLegals()[0]->getRIF();
        }

        $orders[] = $order;
    }

    $response['data'] = $orders;
    return new JsonResponse($response);
}

但是我得到这个错误:

Error: Call to a member function getIdentificationType() on a non-object in /var/www/html/tanane/src/Tanane/BackendBundle/Controller/OrderController.php line 115

也许我的映射是错误的，因为我应该在 Person 和 NaturalPerson 之间有 OneToOne(这听起来不符合我的逻辑，如 DER 所示)或者可能不是，但我不知道如何只获取一条记录的相关属性，我阅读了文档 here还有here但是他们没有谈论这部分或者我没有看到它，有什么建议吗？想法？提示？

尝试使用Repositories和DQL解决问题

我正在 Repository 类中构建一个函数来获取数据，而不是像我的问题那样变得复杂，所以我这样做了:

public function getOrders($person_type = 1)
{
    $qb = $this->getEntityManager()->createQueryBuilder();

    $qb
            ->select('ord.*, ps.*')
            ->from("FrontendBundle:Orders", "ord")
            ->join('FrontendBUndle:Person', 'ps', 'WITH', 'ps.id = ord.person_id')
            ->orderBy('ord.created', 'DESC');

    if ($person_type == 1)
    {
        $qb
                ->select('np.*')
                ->join('FrontendBundle:NaturalPerson', 'np', 'WITH', 'ps.id = np.person'); // Join NaturalPerson table
    }
    elseif ($person_type == 2)
    {
        $qb
                ->select('lp.*')
                ->join('FrontendBundle:LegalPerson', 'lp', 'WITH', 'ps.id = lp.person'); // Join NaturalPerson table
    }

    return $qb->getQuery()->getResult();
}

我还没有测试过，所以它可能行不通，但是，如果我的想法是获取两个表的额外信息，那么我使用这个 DQL 我如何传递 $person_type哪个在 Person 表中？这有点复杂，至少对我来说是这样

运行原始查询以查看列是否为 NULL

我构建这个简单的查询只是为了测试结果是否为 NULL:

SELECT
    ord.id,
    ord.person_id as ord_person_id,
  ord.nickname,
  ps.id,
  ps.description,
  np.person_id as natural_person_id,
  np.identification_type,
    np.ci
FROM
    orders ord
LEFT JOIN person ps ON ord.person_id = ps.id
LEFT JOIN natural_person np ON np.person_id = ps.id
WHERE
    ps.person_type = 1;

这是什么查询返回:

所以那里没有 NULL 列

用于创建新订单的 CRUD

// Set Person entity
$entityPerson = new Person();
$person_type === 1 ? $entityPerson->setDescription($orders['nat']['person']['description']) : $entityPerson->setDescription($orders['leg']['person']['description']);
$person_type === 1 ? $entityPerson->setContactPerson($orders['nat']['person']['contact_person']) : $entityPerson->setContactPerson($orders['leg']['person']['contact_person']);
$entityPerson->setPersonType($person_type);

$em->persist($entityPerson);
$em->flush();

...

if ($person_type === 1)
{
    // Set NaturalPerson entity
    $entityNatural = new NaturalPerson();
    $entityNatural->setIdentificationType($orders['nat']['identification_type']);
    $entityNatural->setCI($orders['nat']['ci']);

    $em->persist($entityNatural);
    $em->flush();
}
elseif ($person_type === 2)
{
    // Set LegalPerson entity
    $entityLegal = new LegalPerson();
    $entityLegal->setIdentificationType($orders['leg']['identification_type']);
    $entityLegal->setRIF($orders['leg']['rif']);

    $em->persist($entityLegal);
    $em->flush();
}

Answer 1

因为 LegalPerson 和 NaturalPerson 是 Person 的特化，所以我建议使用 Doctrine 所谓的类表继承 (documentation)。

你会:

Person.php

/**
 * @ORM\Table(name="person")
 * @ORM\Entity
 * @ORM\InheritanceType("JOINED")
 * @ORM\DiscriminatorColumn(name="discr", type="string")
 * @ORM\DiscriminatorMap({
 *     "natural" = "NaturalPerson",
 *     "legal" = "LegalPerson",
 * })
 */
class Person {
    /**
     * @ORM\OneToMany(targetEntity="Orders", mappedBy="person")
     * */
    private $orders;

    public function __construct()
    {
        $this->orders = new ArrayCollection();
    }

    public function getOrders()
    {
        return $this->orders;
    }

}

NaturalPerson.php

/**
 * @ORM\Table(name="natural_person")
 * @ORM\Entity
 */
class NaturalPerson extends Person {
    /**
     * @ORM\Column(name="identification_type", type="ci_type", nullable=false)
     * @DoctrineAssert\Enum(entity="Tanane\FrontendBundle\DBAL\Types\CIType")
     */
    protected $identification_type;

    /**
     * @ORM\Column(name="ci", type="integer", nullable=false)
     */
    protected $ci;

    public function setIdentificationType($identification_type)
    {
        $this->identification_type = $identification_type;
        return $this;
    }

    public function getIdentificationType()
    {
        return $this->identification_type;
    }

    public function setCI($ci)
    {
        $this->ci = $ci;
        return $this;
    }

    public function getCI()
    {
        return $this->ci;
    }
}

Order.php 保持不变。

如您所见，现在 NaturalPerson 和 LegalPerson 都扩展了 Person。由于您更改了实体定义，因此您必须更新数据库架构。

现在，在您的 Controller 中，您只需执行以下操作:

foreach ($entities as $entity)
{
    $person = $entity->getPerson();

    $order = array();
    $order[] = $entity->getNickname();

    if ($person instanceof NaturalPerson)
    {
        $order[] = $person->getIdentificationType() . $person->getCI();
    }
    else // it has to be LegalPerson
    {
        $order[] = $person->getIdentificationType() . $person->getRIF();
    }

    $orders[] = $order;
}

不要忘记为 NaturalPerson 添加 use 语句!

通过这种方式，您只能使用 NaturalPerson 或 LegalPerson 的实例。我相信您可以进一步改进这一点。

最后，您必须为此更改您的 CRUD。您不再直接使用 Person(实际上，它应该是 abstract)，所以现在您需要处理 NaturalPerson 和分别为 LegalPerson。每个都有它的类型、 Controller 、 View 等。

您的代码现在看起来像这样:

if ($person_type === 1)
{
    $entityPerson = new NaturalPerson();
    $entityPerson->setDescription($orders['nat']['person']['description']);
    $entityPerson->setContactPerson($orders['nat']['person']['contact_person']);
    $entityPerson->setIdentificationType($orders['nat']['identification_type']);
    $entityPerson->setCI($orders['nat']['ci']);

    $em->persist($entityPerson);
    $em->flush();
}
elseif ($person_type === 2)
{
    $entityPerson = new LegalPerson();
    $entityPerson->setDescription($orders['leg']['person']['description']);
    $entityPerson->setContactPerson($orders['leg']['person']['contact_person']);
    $entityPerson->setIdentificationType($orders['leg']['identification_type']);
    $entityPerson->setRIF($orders['leg']['rif']);

    $em->persist($entityPerson);
    $em->flush();
}