具有API平台Symfony的继承实体

问题描述 投票:2回答:1

当我有一个从另一个实体继承的实体时,平台API(https://api-platform.com/)出现问题。例如,从用户实体继承的工作者实体。当我转到Platform API文档时,所有Worker属性都出现在User类中


比解释更好,是一个模式。这是我的两个实体以及有问题的文档的结果

/**
* User
*
* @ORM\Table(name="user")
* @ORM\InheritanceType("SINGLE_TABLE")
* @ORM\DiscriminatorColumn(name="discr", type="string")
* @ORM\DiscriminatorMap({"user" = "User", "worker" = "Worker"})
* @ApiResource
*
*/
class User
{
/**
 * @var int
 *
 * @ORM\Column(name="id", type="integer")
 * @ORM\Id
 * @ORM\GeneratedValue(strategy="AUTO")
 */
protected $id;

/**
 * @var string
 *
 * @ORM\Column(name="lastname", type="string", length=255)
 */
protected $lastname;

/**
 * @var string
 *
 * @ORM\Column(name="firstname", type="string", length=255)
 */
protected $firstname;


/**
 * Get id
 *
 * @return int
 */
public function getId()
{
    return $this->id;
}

/**
 * Set lastname
 *
 * @param string $lastname
 * @return User
 */
public function setLastname($lastname)
{
    $this->lastname = $lastname;

    return $this;
}

/**
 * Get lastname
 *
 * @return string
 */
public function getLastname()
{
    return $this->lastname;
}

/**
 * Set firstname
 *
 * @param string $firstname
 * @return User
 */
public function setFirstname($firstname)
{
    $this->firstname = $firstname;

    return $this;
}

/**
 * Get firstname
 *
 * @return string
 */
public function getFirstname()
{
    return $this->firstname;
}
}


/**
* Worker
*
* @ApiResource
*/
class Worker extends User
{

/**
 * @var \DateTime
 *
 * @ORM\Column(name="birthday", type="datetime")
 * @Assert\NotNull()
 * @Assert\DateTime()
 */
private $birthday;

/**
 * @var string
 *
 * @ORM\Column(name="mobilePhone", type="string", length=255, nullable=true)
 */
private $mobilePhone;

/**
 * @return \DateTime
 */
public function getBirthday(): \DateTime
{
    return $this->birthday;
}

/**
 * @param \DateTime $birthday
 */
public function setBirthday(\DateTime $birthday)
{
    $this->birthday = $birthday;
}

/**
 * @return string
 */
public function getMobilePhone(): string
{
    return ($this->mobilePhone == null) ? '' : $this->mobilePhone;
}

/**
 * @param string $mobilePhone
 */
public function setMobilePhone(string $mobilePhone)
{
    $this->mobilePhone = $mobilePhone;
}

}

这是问题。我们看到Worker子类的属性出现在User类的模型中。而且,当我测试发送POST方法时,它将返回一个包含Worker实体属性的User实体

<< img src =“ https://image.soinside.com/eyJ1cmwiOiAiaHR0cHM6Ly9pLnN0YWNrLmltZ3VyLmNvbS8wS0RFZy5wbmcifQ==” alt =“查看问题的结果”>

symfony api-platform.com
1个回答
0
投票

您的序列化需要进一步深入,以支持在响应中被序列化的worker属性。我不确定如何具体使其适合您的实施。

使用JMS Serialization bundle,看起来像这样:

/**
 * Get the Birthday..
 *
 * @MaxDepth(2)
 * @return \DateTime
 */
public function getBirthday(): \DateTime
{
    return $this->birthday;
}
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