symfony - 与api平台的嵌入关系

Question

我对 Api 平台有疑问。 (https://api-platform.com)我有两个实体。问题和答案。我想要进行一次 POST 调用来创建一个只有一个答案的问题。我显示我的实体。

namespace App\Entity;

use ApiPlatform\Core\Annotation\ApiResource;
use Doctrine\Common\Collections\ArrayCollection;
use Doctrine\ORM\Mapping as ORM;
use Symfony\Component\Validator\Constraints as Assert;
use Symfony\Component\Serializer\Annotation\Groups;

/**
 * @ApiResource(
 *     normalizationContext={"groups"={"question"}},
 *     denormalizationContext={"groups"={"question"}})
 * @ORM\Entity
 */
class Question
{
    /**
     * @Groups({"question"})
     * @ORM\Id
     * @ORM\GeneratedValue
     * @ORM\Column(type="integer")
     */
    private $id;

    /**
     * @Groups({"question"})
     * @ORM\Column
     * @Assert\NotBlank
     */
    public $name = '';

    /**
     * @Groups({"question"})
     * @ORM\OneToMany(targetEntity="Answer", mappedBy="question", cascade={"persist"})
     */
    private $answers;

    public function getAnswers()
    {
        return $this->answers;
    }

    public function setAnswers($answers): void
    {
        $this->answers = $answers;
    }


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

    public function getName(): string
    {
        return $this->name;
    }

    public function setName(string $name): void
    {
        $this->name = $name;
    }

    public function getId(): int
    {
        return $this->id;
    }
}

和一个应答实体

namespace App\Entity;

use ApiPlatform\Core\Annotation\ApiResource;
use Doctrine\ORM\Mapping as ORM;
use Symfony\Component\Validator\Constraints as Assert;
use Symfony\Component\Serializer\Annotation\Groups;

/**
 *
 * @ApiResource
 * @ORM\Entity
 */
class Answer
{
    /**
     * @Groups({"question"})
     * @ORM\Id
     * @ORM\Column(type="guid")
     */
    public $id;

    /**
     * @Groups({"question"})
     * @ORM\Column
     * @Assert\NotBlank
     */
    public $name = '';

    /**
     * @ORM\ManyToOne(targetEntity="Question", inversedBy="answers")
     * @ORM\JoinColumn(name="question_id", referencedColumnName="id")
     */
    public $question;

    public function getQuestion()
    {
        return $this->question;
    }

    public function setQuestion($question): void
    {
        $this->question = $question;
    }

    public function getName(): string
    {
        return $this->name;
    }

    public function setName(string $name): void
    {
        $this->name = $name;
    }

    public function getId(): string
    {
        return $this->id;
    }

    public function __toString()
    {
        return $this->getName();
    }
}

现在我可以从 nelmio 的仪表板创建一个问题并生成答案。但在数据库中，我的答案并没有保存与问题的关系。

{
  "name": "my new question number 1",
  "answers": [
    {
          "id": "ddb66b71-5523-4158-9aa3-2691cae9d473",
          "name": "my answer 1 to question number 1"
    }
  ]
}

另一个问题是...我已经通过 guid 更改了我的答案 id，因为当我创建并回答没有 id 的问题时，我收到错误。我可以创建问题和答案而不指定 ID 吗？

提前致谢

Answer 1

嗯，我在您的实体中看到一些错误...首先，由于问题和答案实体之间的关系是 OneToMany，您的 Qestion 实体应该实现此功能，即:

use ApiPlatform\Core\Annotation\ApiProperty;

//..... the rest of your code

/**
 * @ApiProperty(
 *    readableLink=true
 *    writableLink=true
 * )
 * @Groups({"question"})
 * @ORM\OneToMany(targetEntity="Answer", mappedBy="question", cascade={"persist"})
 */
private $answers;


public function __construct()
{
     //....

      $this->answers = new ArrayCollection();

     //...
}

public function addAnswer(Answer $answer): self
{
     if (!$this->answers->contains($answer)) {
           $this->answers[] = $answer;
           $answer->setQuestion($this) 
     }

     return $this;
}

public function removeAnswer(Answer $answer): self
{
    if ($this->answers->contains($answer)) {
        $this->answers->removeElement($answer);
    }
    return $this;
}

命令

PHP bin/console make:entity

允许您在关系类型的实体中创建一个字段，并且只需按照说明操作即可为您创建这些方法(创建两个实体后，使用更新问题实体的命令...)

ReadLink ApiProperty 注释用于查看 GET 请求上的嵌入对象，如果您使用序列化组，则相同，如果将其设置为 false，则响应将如下所示:

{
  "name": "my new question number 1",
  "answers": [
    "/api/answers/1",
    "/api/answers/2",
    ....
  ]
}

它用于使响应更小(除其他外)...而 writableLink 用于允许像这样的 POST 请求(有关更多信息，请参阅此示例 here ):

{
  "name": "my new question number 1",
  "answers": [
    {
          "id": "ddb66b71-5523-4158-9aa3-2691cae9d473",
          "name": "my answer 1 to question number 1"
    }
  ]
}

当然，在每个实体中使用相应的序列化组......在 ApiPlatform 中，嵌入对象通过 setter 和 getters 方法进行持久化，还为 OneToMany 关系添加和删除方法，ORM 完成其余工作。让我知道这是否有帮助。干杯!