3

In a Sonata admin form, I'd like to display all images related to the current object. Don't need to edit them. I managed to get this working with an entity field type and its property option to select the image URL as label so I can call all pictures in the template :

->add('image', 'entity', array(
    'class' => 'Hop\HopBundle\Entity\Image',
    'property' => 'image_url',
    'query_builder' => function($er) use ($object) {
        return $er->createQueryBuilder('i')
        ->where('i.object = :object')
        ->setParameter('object', $object)
        ;
    },
    'expanded' => true,
))

.

{% for child in form %}
    <img src="{{ child.vars.label }}" alt="{{ child.vars.label }}" />
{% endfor %}

It's because it seems that the entity type only give label and one value to the template.

But I'd like to get also image width, descriptions, date, ... In other word : how can we get each entire related image object in the Twig template ?

Thanks.

4

3 回答 3

6

所以我用谷歌搜索了很多，这对我有用（Symfony 2.2）
更新为@flu在评论中建议

{% for key, child in form %}
<div>
    {% set entity = form.vars.choices[key].data %}
    {{ form_widget(child) }}
    {{ form_label(child) }}
</div>
{% endfor %}

并且entity是子对象。form.vars.data从 Symfony 2.3 开始，如果我理解正确，将会有： https ://github.com/symfony/symfony/pull/5023

于 2013-05-03T12:53:33.910 回答
4

我花了一整天的时间来解决这个问题。

首先，我创建了一个扩展 ChoiceType 的新 FormType：

class NomineesType extends AbstractType
{
protected $em;

public function __construct(EntityManager $em) {
    $this->em = $em;
}

public function setDefaultOptions(OptionsResolverInterface $resolver)
{
    $submissionRepository = $this->em->getRepository('BundleName:Submission');
    $choices = $submissionRepository->findBy(array(
        'nominee' => true
    ));
    $resolver->setDefaults(array(
        'choices' => $choices
    ));
}

public function getName()
{
    return 'nominees';
}

public function getParent()
{
    return 'choice';
}
}

比我必须注册的服务：

services:
form.type.nominees:
    class: Bundle\Form\Type\NomineesType
    arguments:
        entityManager: "@doctrine.orm.entity_manager"
    tags:
        - { name: form.type, alias: nominees }

之后，我将新类型添加到我的表单中：

public function buildForm(FormBuilderInterface $builder, array $options)
{
    $game = $this->game;
    $builder->add('submission', 'nominees', array(
        'expanded' => true,
        'multiple' => false
    ));
    $builder->add('email');
}

现在我可以在我的 twig-template 中获取所有属性：

{% block nominees_widget %}
{% spaceless %}
    {% if expanded %}
        <ul {{ block('widget_container_attributes') }}>
        {% for child in form %}
            <li>
                {{ child.get('form').get('label').firstname }}
                {{ child.get('form').get('label').lastname }}
                {{ form_widget(child) }}
                {{ form_label(child) }}
            </li>
        {% endfor %}
        </ul>
    {% else %}
        {# just let the choice widget render the select tag #}
        {{ block('choice_widget') }}
    {% endif %}
{% endspaceless %}
{% endblock %}

因为选择列表只返回整数值，所以我创建了一个新的数据转换器来将选择的 id 转换为实体：

class IntToEntityTransformer implements DataTransformerInterface
{
/**
 * @var ObjectManager
 */
private $om;

/**
 * @param ObjectManager $om
 */
public function __construct(ObjectManager $om)
{
    $this->om = $om;
}

/**
 * Transforms an object (issue) to a string (number).
 *
 * @param  Issue|null $issue
 * @return string
 */
public function transform($issue)
{
    if (null === $issue) {
        return "";
    }

    return $issue->getNumber();
}

/**
 * Transforms a string (number) to an object (issue).
 *
 * @param  string $number
 * @return Issue|null
 * @throws TransformationFailedException if object (issue) is not found.
 */
public function reverseTransform($number)
{
    if (!$number) {
        return null;
    }

    $issue = $this->om
        ->getRepository('BundleName:Submission')
        ->findOneBy(array('id' => $number))
    ;

    if (null === $issue) {
        throw new TransformationFailedException(sprintf(
            'An Submission with number "%s" does not exist!',
            $number
        ));
    }

    return $issue;
}
}

此数据转换器在构建表单方法中初始化：

public function buildForm(FormBuilderInterface $builder, array $options)
{
    $transformer = new IntToEntityTransformer($this->em);
    $builder->add($builder->create('submission', 'nominees', array(
        'expanded' => true,
        'multiple' => false
    ))->addModelTransformer($transformer));
    $builder->add('email');
}
于 2012-09-08T13:41:59.247 回答
0

我一直在用 Symfony 2.3 解决这个问题。不知道它在 2.1 中是如何工作的。

在我的情况下，我有一个实体选择（扩展和多个），我需要在每个选择的标签旁边添加实体的“描述”字段。

我解决如下：

1- 创建了一个自定义类型，其父类型为 EntityType，名称为“xxxxxxx_form_reparacionnormalizada”

class ReparacionNormalizadaType extends AbstractType
{
    public function setDefaultOptions(OptionsResolverInterface $resolver)
    {
        $resolver->setDefaults(array(
            'class'        => 'xxxxxxxxx\Entity\TipoReparacion',
        ));
    }

    public function getName()
    {
        return 'xxxxxxx_form_reparacionnormalizada';
    }

    public function getParent()
    {
        return 'entity';
    }
}

2-像往常一样将表单类型注册为服务

3-创建了一个自定义模板，如下所示：

{% block xxxxxxx_form_reparacionnormalizada_widget %}
<ul {{ block('widget_container_attributes_choice_widget') }}>

    {% for id, child in form.children %}

    <li>
        {{ form_label(child, child.vars.label|default(null), { 'in_list_checkbox' : true, 'widget' : form_widget(child) } ) }}

        <span>{{ choices[id].data.descripcion }}</span>

    </li>
    {% endfor %}
</ul>
{% endblock %}

4-在您的奏鸣曲管理员中，您使用自定义类型而不是'entity'

希望能帮助到你

问候

哈维尔

于 2013-07-24T15:11:48.133 回答