Symfony 西蒙尼。如何构建表单？

如何实现接受以下JSON对象的表单？ 我没有定义模型，因为我不需要它们。数据被发送

{
  "type_of_error": "logic error",
  "severity": "normal",
  "what_did_you_do": "something",
  "what_happened": "blue screen",
  "which_result": "full satisfaction",
  "requests": [
    {
      "url": "/api/v1/agent/statistic"
    }
  ]
}

我已经开始构建表单：

class ErrorReportFormType extends AbstractType
{
    public function buildForm(FormBuilderInterface $builder, array $options)
    {
        $builder
            ->add('type_of_error', TextType::class, [
                'empty_data' => '',
                'constraints' => [
                    new NotBlank()
                ]
            ])
            ->add('severity', TextType::class, [
                'empty_data' => '',
                'constraints' => [
                    new NotBlank()
                ]
            ])
            ->add('what_did_you_do', TextType::class, [
                'empty_data' => '',
                'constraints' => [
                    new NotBlank()
                ]
            ])
            ->add('what_happened', TextType::class, [
                'empty_data' => '',
                'constraints' => [
                    new NotBlank(),
                ]
            ])
            ->add('which_result', TextType::class, [
                'empty_data' => '',
                'constraints' => [
                    new NotBlank(),
                ]
            ])
            ->add('requests', CollectionType::class, [
                'entry_type'   => RequestsFormType::class,
            ])
        ;
    }

我已经像这样定义了RequestsFormType。这里怎么了？看起来，好像RequestsFormType不会被接受

class RequestsFormType extends AbstractType
{

    public function buildForm(FormBuilderInterface $builder, array $options)
    {
        $builder
            ->add('url', TextType::class, [
                'empty_data' => '',
                'constraints' => [
                    new NotBlank()
                ]
            ]);
    }

    public function getBlockPrefix()
    {
        return '';
    }
}

---

也许可以尝试使用CollectionType

请看这里：

您的自定义类型：

class YourCustomType extends AbstractType
{
    public function buildForm(FormBuilderInterface $builder, array $options)
    {
        $builder->add('url', TextType::class, []);
        $builder->add('method', TextType::class, []);
        $builder->add('timestamp', DateTimeType::class, []);
    }

    public function configureOptions(OptionsResolver $resolver)
    {
        $resolver->setDefaults([
            'data_class' => Requests::class,
        ]);
    }
}

---

如何定义url、方法和时间戳？在Collectiontype中给出参数：

'entry\u type'=>YourCustomType:：class，

创建一个自定义表单并在生成器中添加url、方法和时间戳我应该定义一个额外的类吗？没有别的解决办法吗？以及如何在buildForm方法中创建自定义表单？是的，您应该创建一个自定义类型，如RequestsType，并在FormBuilder中添加参数：url、方法和时间戳。我不知道是否还有其他解决办法。这个应该适合你的需要好的，它看起来像是另一种形式中的一种形式。

class YourCustomType extends AbstractType
{
    public function buildForm(FormBuilderInterface $builder, array $options)
    {
        $builder->add('url', TextType::class, []);
        $builder->add('method', TextType::class, []);
        $builder->add('timestamp', DateTimeType::class, []);
    }

    public function configureOptions(OptionsResolver $resolver)
    {
        $resolver->setDefaults([
            'data_class' => Requests::class,
        ]);
    }
}