Question

我需要访问表单对象中的属性。问题是，我想访问的属性不是在表单中呈现，也不是在contractType类中声明的。

    class ContractType extends AbstractType
    {
        /**
         * @param FormBuilderInterface $builder
         * @param array $options
         */
        public function buildForm(FormBuilderInterface $builder, array $options)
        {
            $builder
                ->add('rcode1', new TextType(), array('label' => 'rcode 1'))
                ->add('rcode2', new TextType(), array('label' => 'rcode 2'));
        }
...
}

转储表单对象：

array:6 [▼
  "contract" => Contract {#2003 ▼

    - id: null
    - actionCode: "104"
    - productCode: "20106"
    - created: null
    - updated: null
    - resumeId: null
    - rcode1: null
    - rcode2: null
    - downloadId: null
    - businessContractDetails: BusinessContractDetails {#1999 ▶}
    - privateContractDetails: null
    - company: Company {#2000 ▶}
    - persons: ArrayCollection {#1998 ▶}
  }

"businessContractDetails" => BusinessContractDetails {#1999 ▶}

"company" => Company {#2000 ▶}

"contactPerson" => ContactPerson {#1987 ▶}

"landlord" => Landlord {#1993 ▶}

"businessRealEstate" => BusinessRealEstate {#1994 ▶}

]

合同实体的属性是rcode1和rcode2。但我需要访问downloadID。

我试过

$form->get('contract')->get('downloadId')->getData();

并收到以下错误消息： Child＆＃34; downloadId＆＃34;不存在。

有什么建议吗？提前谢谢！

Answer 1

尝试

$form->get('contract')->getData()->getDownloadId();

请注意我假设您的getDownloadId课程中有一个getter Contract

Answer 2

您的控制器不应通过表单访问实体的值。它会起作用，但这是一种不好的做法，例如，如果表单类型contract的名称会发生变化。

在handleRequest该实体填充表单中的数据之后，您可以依赖您创建表单的实体。所以：

# Controller POST action
$entity = new Contract(); // Or retrieve it from database: $em->find(Contract, $id);
$form = $this->createForm(FormType::class, $entity)->handleRequest($request);
echo $entity->getDownloadId();

Symfony 2.7：在表单对象中访问未呈现的属性

2 个答案: