Symfony 类型为“的预期参数”;字符串“&引用;“应用\实体”;鉴于

Symfony 类型为“的预期参数”;字符串“&引用;“应用\实体”;鉴于,symfony,doctrine-orm,orm,one-to-many,symfony4,Symfony,Doctrine Orm,Orm,One To Many,Symfony4,我试图插入一个又一个关系的记录,但出现了错误 Expected argument of type "string", "App\Entity\Question" given. 我有下一个实体问题和答案 class Question { /** * @ORM\Id() * @ORM\GeneratedValue() * @ORM\Column(type="integer") */ private $id; /** * @ORM\Colum

我试图插入一个又一个关系的记录,但出现了错误

Expected argument of type "string", "App\Entity\Question" given.
我有下一个实体
问题
答案

class Question
{
    /**
     * @ORM\Id()
     * @ORM\GeneratedValue()
     * @ORM\Column(type="integer")
     */
    private $id;
/**
 * @ORM\Column(type="text")
 */
private $title;

/**
 * @ORM\OneToMany(targetEntity="App\Entity\Answer", mappedBy="question", orphanRemoval=true)
 */
private $answers;

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

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

public function getTitle(): ?string
{
    return $this->title;
}

public function setTitle(string $title): self
{
    $this->title = $title;

    return $this;
}

/**
 * @return Collection|Answer[]
 */
public function getAnswers(): Collection
{
    return $this->answers;
}

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);
        if ($answer->getQuestion() === $this) {
            $answer->setQuestion(null);
        }
    }

    return $this;
}
}
实体
答案

class Answer
{
    /**
     * @ORM\Id()
     * @ORM\GeneratedValue()
     * @ORM\Column(type="integer")
     */
    private $id;
/**
 * @ORM\Column(type="text")
 */
private $text;

/**
 * @ORM\Column(type="boolean")
 */
private $is_correct;

/**
 * @ORM\ManyToOne(targetEntity="App\Entity\Question", inversedBy="answers")
 * @ORM\JoinColumn(nullable=false)
 */
private $question;

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

public function getText(): ?string
{
    return $this->text;
}

public function setText(string $text): self
{
    $this->text = $text;

    return $this;
}

public function getIsCorrect(): ?bool
{
    return $this->is_correct;
}

public function setIsCorrect(bool $is_correct): self
{
    $this->is_correct = $is_correct;

    return $this;
}

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

public function setQuestion(?Question $question): self
{
    $this->question = $question;

    return $this;
}
}

我的表格

public function buildForm(FormBuilderInterface $builder, array $options)
{
    $builder
        ->add('title', EntityType::class, array(
            'class' => Question::class,
            'choice_label' => 'title',
            'label' => 'Question'
        ));
    $builder
        ->add('answers', CollectionType::class, array(
        'entry_type' => AnswerType::class,
        'entry_options' => array('label' => false),
        'allow_add' => true,
        'by_reference'  => false,
));
    $builder
        ->add('create', SubmitType::class, ['label' => 'Add', 'attr' => ['class' => 'btn btn-primary']]);


}

public function configureOptions(OptionsResolver $resolver)
{
    $resolver->setDefaults([
        'data_class' => Question::class
    ]);
}
我的控制器碎片

 $question = new Question();
 $answer = new Answer();
 $question->addAnswer($answer);
 $form1 = $this->createForm(QuestionAnswerType::class, $question);
 $form1->handleRequest($request);
    if ($form->isSubmitted() && $form->isValid()) {
        $em = $this->getDoctrine()->getManager();
        $em->persist($question);
        $em->flush();
    }
错误指针位于下一行

 $form1->handleRequest($request);
我知道我的控制器有问题,但我不知道如何解决


我不明白如何正确地插入关系记录。您能帮助我吗?

我认为您看到此错误的原因是因为您在
问题
类中将
标题
字段定义为文本类型(
@ORM\Column(type=“Text”)

但是,在表单中,您将表单字段
title
定义为
EntityType
,这就是我认为您看到此错误的原因

您可以通过更改
问题中标题字段的数据库映射来解决此问题
您可以将表单更改为使用
文本类型而不是
实体类型


希望这有帮助

您必须在两个地方进行更改

1) 首先改为“问题”类

2) 在表单类中,将“EntityType::class”替换为“TextType::class”,并从标题中删除“class”和“choice_label”属性

public function buildForm(FormBuilderInterface $builder, array $options)
{
    $builder
        ->add('title', TextType::class, array(
            'label' => 'Question'
        ));
    ..... your other code ...   
}

我把实体的类型改成了“字符串”,但仍然有错误。你们应该看看Kapil的答案,我认为这是正确的方法。我明白了。但我需要使用EntityType。EntityType仅在您与其他表有关系时使用。这里的“title”字段与任何一个字段都没有关联,它是一个简单的文本类型。我在表question和answer之间有很多关系。True。为此,您使用了“私人$answers;”。“标题”列与其他表没有关联。它是独立的和列的。希望你得到它。:)我必须使用什么来代替EntityType的标题?
public function buildForm(FormBuilderInterface $builder, array $options)
{
    $builder
        ->add('title', TextType::class, array(
            'label' => 'Question'
        ));
    ..... your other code ...   
}