在symfony中以嵌套形式symfony（孩子的孩子）访问按钮

问题描述：

This is form i use:

$form = $this->createForm(new NewsType(), $news)
            ->add('edit', SubmitType::class, array('label' => 'edit'))
            ->add('delete', SubmitType::class, array('label' => 'delete'))
            ->add('comments', CollectionType::class, array('entry_type'   => CommentType::class));

CommentType:

 public function buildForm(FormBuilderInterface $builder, array $options)
{
    $builder
        ->add('author', TextType::class)
        ->add('text', TextType::class)
        ->add('remove', SubmitType::class);
}

public function configureOptions(OptionsResolver $resolver)
{
    $resolver->setDefaults(array(
        'data_class' => 'AppBundle\Entity\Comment'));
}

Is it possible to access remove button from CommentType so when its clicked to delete comment entry. Everything is mapped properly, i can see comment objects displayed on my page, but when i use $form->get('remove') i get "Child "remove" does not exist." Is it even possible to do this way?

答

You need to access a grand grand child doing:

foreach ($form->get('comments') as $entry) {
    $toRemove = $entry->get('remove')-isClicked();
    // handle it ...
}

But to submit it separately you must ensure that your building the "complete" child form in your view:

{{ form_start(form) }}
{% for child in form %}
    {% if 'news_comments' == child.vars['full_name'] %}
        {{ form_start(child) }}
        {{ form_row(child) }}
        {{ form_end(child) }}
    {% else %}
        {{ form_row(child) }}
    {% endif %}
{% endfor %}
{{ form_end(form) %}

Aside note:

be careful, you seem to use symfony 2.8 and to have updated the FQCN for the form types, but it's needed too for creating the form:

$form = $this->createForm(NewsType::class, $news)

在symfony中以嵌套形式symfony（孩子的孩子）访问按钮

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