Symfony 形式 - 类型为“字符串"的预期参数在属性路径中给出

问题描述

我收到此错误消息:类型为App\Entity\Artist 或 null"的预期参数，在属性路径artist"中给出的string".

I get this error message:Expected argument of type "App\Entity\Artist or null", "string" given at property path "artist".

这是我在 symfony 中的表单代码

This is the code of my form in symfony

$builder
        ->add('date', DateType::class, [
            'widget' => 'single_text'
        ])
        ->add('artist', TextType::class)
        ->add('City',TextType::class)
        ;

这是模板:

{{ form_start(form)}}
    {{ form_widget(form)}}
        <button class="btn btn-secondary">{{ button|default('Enregistrer')}}</button>
{{ form_end(form)}}

实体

class Artist
{
/**
 * @ORM\Id()
 * @ORM\GeneratedValue()
 * @ORM\Column(type="integer")
 */
private $id;
/**
 * @ORM\Column(type="string", length=45)
 */
private $name;
/**
 * @ORM\OneToMany(targetEntity="App\Entity\Event", mappedBy="artist", cascade={"persist","remove"})
 */
private $events;
public function __construct()
{
    $this->events = new ArrayCollection();
}
public function getId(): ?int
{
    return $this->id;
}
public function getName(): ?string
{
    return $this->name;
}
public function setName(?string $name): self
{
    $this->name = $name;
    return $this;
}
/**
 * @return Collection|Event[]
 */
public function getEvents(): Collection
{
    return $this->events;
}
public function addEvent(Event $event): self
{
    if (!$this->events->contains($event)) {
        $this->events[] = $event;
        $event->setArtist($this);
    }
    return $this;
}
public function removeEvent(Event $event): self
{
    if ($this->events->contains($event)) {
        $this->events->removeElement($event);
        // set the owning side to null (unless already changed)
        if ($event->getArtist() === $this) {
            $event->setArtist(null);
        }
    }
    return $this;
}
public function __toString()
{
    return $this->name;
}
}

这是错误:

Uncaught PHP Exception Symfony\Component\PropertyAccess\Exception\InvalidArgumentException: "Expected argument of type "App\Entity\Artist or null", "string" given at property path "artist"." at D:\wamp64\www\app_music_events\vendor\symfony\property-access\PropertyAccessor.php line 173

{
    "exception": {}
}

这是控制器

/**
 * Method for editing an event
 *
 * @Route("/admin/edit/{id}", name="admin.edit", methods="GET|POST")
 * @param Request $request
 * @return \Symfony\Component\HttpFoundation\Response
 */
public function edit(Event $event, Request $request): Response
{
    $form = $this->createForm(EventType::class, $event);
    $form->handleRequest($request);

    if ($form->isSubmitted() && $form->isValid()) {
        $this->em->flush();
        $this->addFlash('success', 'Le concert a été modifié avec succès !');
        return $this->redirectToRoute('admin.index');
    }

    return $this->render('admin/edit.html.twig', [
            'event' => $event,
            'form' => $form->createView()
        ]
    );
}

我不明白.当我没有在表单中指定 Textype 时，我没有这个问题.

I don't understand.I do not have this problem when I do not specify Textype in the form.

有什么问题?谢谢

推荐答案

问题是您需要将 Artist 实体与您的事件一起持久化.Symfony 需要一个 Artist 实体类型而不是一个字符串.尝试使用 EntityType::class 而不是 TextType::class

The problem is that you need to persist an Artist entity along with your Event. Symfony expects an Artist entity type and not a string. Try with EntityType::class instead of TextType::class

->add('artist', EntityType::class, [
   'class' => Artist::class
]);

有关 EntityType 字段的更多信息，请检查文档

For more information on the EntityType field, please check the documentation

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