Php Symfony FormType-在多个选项上选择
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: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;
// 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) }}