Php Symfony FormType-在多个选项上选择

Php Symfony FormType-在多个选项上选择,php,forms,symfony,doctrine-orm,Php,Forms,Symfony,Doctrine Orm,我尝试为我的Symfony应用程序创建一个新的FormType。我有一个多通字段: // Grower.php /** * @ORM\ManyToOne(targetEntity="CategoryGrower", inversedBy="growers") * @ORM\JoinColumn(name="category_id", referencedColumnName="id") **/ private $category;

我尝试为我的Symfony应用程序创建一个新的FormType。我有一个多通字段:

    // Grower.php
    /**
     * @ORM\ManyToOne(targetEntity="CategoryGrower", inversedBy="growers")
     * @ORM\JoinColumn(name="category_id", referencedColumnName="id")
     **/
    private $category;

    // CategoryGrower.php
    /**
     * @ORM\OneToMany(targetEntity="Grower", mappedBy="category")
     **/
    private $growers;

    // GrowerType.php
    $builder
        ->add('name', 'text')
        ->add('category', 'entity', array(
            'class' => 'FermierMalin\Bundle\GrowerBundle\Entity\CategoryGrower',
            'choice_label' => 'label',
            'group_by' => 'parent',
        ));
但select in HTML代码为空:

<div>                
<label for="fermiermalin_bundle_growerbundle_grower_category" class="required">Category</label>        
<select id="fermiermalin_bundle_growerbundle_grower_category" name="fermiermalin_bundle_growerbundle_grower[category]" required="required">
</select>
</div>
我是这样做的

实体

 /**
 * @ORM\ManyToOne(targetEntity="FamebitBundle\Entity\Price")
 * @ORM\JoinColumn(name="price_id", referencedColumnName="id")
 */
protected $price;
模板

->add('price', EntityType::class, array(
            'class' => 'FamebitBundle\Entity\Price',
            'property' => 'label',
            'choice_value' => 'id',
        ))
细枝视图

{{ form_row(form.price) }}
这对我来说很有用:)

你能给全班(种植者和分类种植者)一个getter/setter和constructor吗?@hubertlocher,
{{ form_row(form.price) }}