访问嵌入式表单数据 symfony

Question

我正在使用 symfony 中的嵌入式表单，但无法从嵌入式表单中获取数据。

这是我的专利表格

namespace AppBundle\Form;

use Symfony\Component\Form\AbstractType;
use Symfony\Component\Form\FormBuilderInterface;
use Symfony\Component\OptionsResolver\OptionsResolver;
use Symfony\Component\Form\Extension\Core\Type\SubmitType;
use Symfony\Component\Form\Extension\Core\Type\TextType;
use Symfony\Component\Form\Extension\Core\Type\EmailType;
use Symfony\Component\Form\Extension\Core\Type\CollectionType;
use AppBundle\Form\SubscriberAddressType;

class SubscriberDetailsType extends AbstractType {
    /**
     * @param FormBuilderInterface $builder
     * @param array                $options
     */
    public function buildForm(FormBuilderInterface $builder, array $options) {

        $builder
            ->add('firstname', TextType::class, [
                'label' => false,
                'required' => true,
                'error_bubbling' => true,
                'attr' => [
                    'placeholder' => 'First Name',
                    'class' => 'form-control'
                    ]])
            ->add('lastname', TextType::class, [
                'label' => false,
                'required' => true,
                'error_bubbling' => true,
                'attr' => array(
                    'placeholder' => 'Last Name',
                    'class' => 'form-control'
                )])
            ->add('emailaddress', EmailType::class, [
                'label' => false,
                'required' => true,
                'error_bubbling' => true,
                'attr' => [
                    'placeholder' => 'Email Address',
                    'pattern'     => '.{2,}',//minlength
                    'class' => 'form-control'
                    ]])
            ->add('subscriberaddress', CollectionType::class, ['entry_type' => SubscriberAddressType::class])
            ->add('submit', SubmitType::class, [
                'label' => 'Sign Up',
                'attr' => [
                    'class' => 'smoothScroll btn btn-danger sb-button'
                    ]
        ])
             ;
    }

    /**
    * @param OptionsResolverInterface $resolver
    */
    public function configureOptions(OptionsResolver $resolver) {
        $resolver->setDefaults(['data_class' => 'AppBundle\Entity\SubscriberDetails']);
    }
    /**
     * @return string
     */
    public function getName() {
        return 'subscriberdetails';
    }
}

这是我的 child 表格

namespace AppBundle\Form;

use Symfony\Component\Form\AbstractType;
use Symfony\Component\Form\FormBuilderInterface;
use Symfony\Component\OptionsResolver\OptionsResolver;
use Symfony\Component\Form\Extension\Core\Type\TextType;

class SubscriberAddressType extends AbstractType {
    /**
     * @param FormBuilderInterface $builder
     * @param array                $options
     */
    public function buildForm(FormBuilderInterface $builder, array $options) {

        $builder
            ->add('address1', TextType::class, [
                'label' => false,
                'required' => true,
                'error_bubbling' => true,
                'attr' => [
                    'placeholder' => 'Address 1',
                    'class' => 'form-control'
                    ]])
            ->add('city', TextType::class, [
                'label' => false,
                'required' => true,
                'error_bubbling' => true,
                'attr' => array(
                    'placeholder' => 'City',
                    'class' => 'form-control'
                )])
             ;
    }

    /**
    * @param OptionsResolverInterface $resolver
    */
    public function configureOptions(OptionsResolver $resolver) {
        $resolver->setDefaults(['data_class' => 'AppBundle\Entity\SubscriberAddress']);
    }
    /**
     * @return string
     */
    public function getName() {
        return 'subscriber';
    }
}

这就是我在我的 Twig 模板中调用嵌入表单的方式

        <div class="form-group ">        
            {% for field in form.subscriberaddress %}
                {{ form_label(field.city) }}
                {{ form_widget(field.city) }}
            {% endfor %}           
        </div>

当我尝试使用下面的表单获取数据时：

            $firstname = $form['firstname'] ->getData();
            $lastname =$form['lastname'] ->getData();
            $email = $form['emailaddress'] ->getData();
            $address = $form['subscriber address']['address1'] ->getData();
            $city =$form['subscriber address']['city'] ->getData();

它在前 3 行以上工作正常，但是当它到达 $address = $form['subscriber address']['address1'] ->getData(); 时with 是一个嵌入式字段，symfony 会报错 child element does not exist.

我已经尝试浏览官方 symfony 食谱以及这些帖子 Post1, Post2 但其中 none 似乎回答了我的问题，即：我如何从表单中的嵌入字段中获取数据？

Answer 1

您的嵌入表单实际上嵌入在 collection 中（不确定这是否是您的意图）因此您需要将其视为 collection/array

如果collection是错误的

您可能想将其更改为

->add('subscriberaddress', SubscriberAddressType::class)

然后 $form->getData() 将起作用。或者，您可以致电 $form->get('subscriberaddress')->getData()

如果你真的想要collection

如果您确实想将其用作 CollectionType 以保留 SubscriberAddressType 的多个副本，则通过

访问数据

foreach ($form->get('subscriberaddress') as $subForm) {
    $subData[] = $subForm->getData();
}

同时考虑将名称 subscriberaddress 更改为 subscriberAddresses 以表明它是 collection

访问嵌入式表单数据 symfony

Access embedded form data symfony

php

symfony-forms

symfony

doctrine-orm

如果collection是错误的

如果你真的想要collection