"类型为"int 或 null"的预期参数，在属性路径中给出的"对象"

Question

我最近在学习Symfony。我遇到一个错误，打电话时找不到一个严肃的答案。

错误如下：未捕获的PHP异常Symfony\Component\PropertyAccess\Exception\InvalidArgumentException："预期的参数类型为"int或null"，"object"，在属性路径"parent_id"处给定。">

我的控制器文件中有以下代码：

public function new(Request $request): Response
{
$category = new Category();
$form = $this->createForm(CategoryType::class, $category);
$form->handleRequest($request);
if ($form->isSubmitted() && $form->isValid()) {
$entityManager = $this->getDoctrine()->getManager();
$entityManager->persist($category);
$entityManager->flush();
return $this->redirectToRoute('category_index');
}
return $this->render('category/new.html.twig', [
'category' => $category,
'form' => $form->createView(),
]);
}

我的FormType文件如下所示：

<?php
namespace AppForm;
use AppEntityCategory;
use SymfonyBridgeDoctrineFormTypeEntityType;
use SymfonyComponentFormAbstractType;
use SymfonyComponentFormFormBuilderInterface;
use SymfonyComponentOptionsResolverOptionsResolver;
class CategoryType extends AbstractType
{
public function buildForm(FormBuilderInterface $builder, array $options)
{
$builder
->add('name', null, [
'attr' => ['class' => 'form-control'],
])
->add('slug', null, [
'attr' => ['class' => 'form-control'],
])
->add('parent_id', EntityType::class, [
'class' => Category::class,
'choice_label' => 'name',
'attr' => ['class' => 'form-control'],
'placeholder' => 'Üst Kategori Seçiniz',
])
->add('title', null, [
'attr' => ['class' => 'form-control']
])
->add('description', null, [
'attr' => ['class' => 'form-control']
])
;
}
public function configureOptions(OptionsResolver $resolver)
{
$resolver->setDefaults([
'data_class' => Category::class,
'required' => false
]);
}
}

我的实体文件的内容在下面

<?php
namespace AppEntity;
use DoctrineORMMapping as ORM;
use SymfonyComponentValidatorConstraints as Assert;
use SymfonyComponentValidatorMappingClassMetadata;
use SymfonyBridgeDoctrineValidatorConstraintsUniqueEntity;
/**
* @ORMEntity(repositoryClass="AppRepositoryCategoryRepository")
*/
class Category
{
/**
* @ORMId()
* @ORMGeneratedValue()
* @ORMColumn(type="integer")
*/
private $id;
/**
* @ORMColumn(type="string", length=255, nullable=true)
*/
private $name;
/**
* @ORMColumn(type="string", length=255, nullable=true)
*/
private $slug;
/**
* @ORMColumn(type="integer", nullable=true)
*/
private $parent_id;
/**
* @ORMColumn(type="string", length=255)
*/
private $title;
/**
* @ORMColumn(type="string", length=255, nullable=true)
*/
private $description;
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;
}
public function getSlug(): ?string
{
return $this->slug;
}
public function setSlug(?string $slug): self
{
$this->slug = $slug;
return $this;
}
public function getParentId(): ?int
{
return $this->parent_id;
}
public function setParentId(?int $parent_id): self
{
$this->parent_id = $parent_id;
return $this;
}
public function getTitle(): ?string
{
return $this->title;
}
public function setTitle(string $title): self
{
$this->title = $title;
return $this;
}
public function getDescription(): ?string
{
return $this->description;
}
public function setDescription(?string $description): self
{
$this->description = $description;
return $this;
}
public function __toString()
{
return $this->name;
}
public static function loadValidatorMetadata(ClassMetadata $metadata)
{
$metadata->addPropertyConstraint('name', new AssertNotBlank());
$metadata->addPropertyConstraint('title', new AssertNotBlank());
$metadata->addPropertyConstraint('description', new AssertNotBlank());
$metadata->addPropertyConstraint('slug', new AssertNotBlank());
$metadata->addConstraint(new UniqueEntity([
'fields' => 'slug',
]));
}
}

如果你能帮我，我将不胜感激。方便。。。

Answer 1

问题是您没有正确配置实体或表单。目前，在您的实体中，parent_id的定义如下：

/**
* @ORMColumn(type="integer", nullable=true)
*/
private $parent_id;

在您的表单中，您期望有另一个Category对象：

->add('parent_id', EntityType::class, [
'class' => Category::class,

您需要将实体中的parent_id设置为一对多关系(其中一个Category可以是另一个的父类别(，或者将该字段设置为常规ChoiceType，并手动提供选项(通过从数据库中获取可能的类别ID(。您将在控制器操作中获得这些id，并将它们作为第三个名为options的参数提供给$this->createForm。