【发布时间】:2020-03-17 10:41:25
【问题描述】:
我已经在脑海中扭转了很长一段时间,但仍然无法找到解决问题的方法。使用 Symfony 4 表单和约束,我无法设置检查以说明在提交包含子表单的表单时,两个字段中的至少一个不能为空。
我有一个 Booking 实体,其中包含一个具有 phoneNumber 属性和 email 属性的 Visitor 实体。我希望能够创建一个具有“访问者”CollectionType 的 Booking(允许我从 BookingType 表单添加访问者)。
我的 BookingType 表单(有点简化):
class BookingType extends AbstractType
{
private $router;
private $translator;
public function __construct(UrlGeneratorInterface $router, TranslatorInterface $translator)
{
$this->router = $router;
$this->translator = $translator;
}
public function buildForm(FormBuilderInterface $builder, array $options)
{
$builder
->add('bookableTimeSlot', EntityType::class, [
'label' => 'entity.booking.bookable-time-slot',
'class' => BookableTimeSlot::class,
'choice_label' => function ($bookableTimeSlot) {
return $bookableTimeSlot->getStartDateTime()->format('d.m.Y h\hi');
}
])
->add('visitors', CollectionType::class, [
'entry_type' => VisitorType::class,
'label' => 'entity.booking.visitors',
'allow_add' => true,
'by_reference' => false,
'entry_options' => ['label' => false]
])
;
}
public function configureOptions(OptionsResolver $resolver)
{
$resolver->setDefaults([
'data_class' => Booking::class,
'user' => User::class,
]);
}
}
我的访客实体(有点简化):
<?php
namespace App\Entity;
use Doctrine\ORM\Mapping as ORM;
use Symfony\Component\Validator\Constraints as Assert;
use Symfony\Component\Validator\Context\ExecutionContextInterface;
/**
* @ORM\Entity(repositoryClass="App\Repository\VisitorRepository")
*/
class Visitor
{
/**
* @ORM\Id()
* @ORM\GeneratedValue()
* @ORM\Column(type="integer")
*/
private $id;
/**
* @ORM\Column(type="string", length=255)
*/
private $firstName;
/**
* @ORM\Column(type="string", length=255)
*/
private $lastName;
/**
* @ORM\Column(type="string", length=45, nullable=true)
*/
private $phone;
/**
* @ORM\ManyToOne(targetEntity="App\Entity\Booking", inversedBy="visitors")
* @ORM\JoinColumn(nullable=false)
*/
private $booking;
/**
* @ORM\Column(type="string", length=255, nullable=true)
*/
private $email;
public function getId(): ?int
{
return $this->id;
}
public function getFirstName(): ?string
{
return $this->firstName;
}
public function setFirstName(string $firstName): self
{
$this->firstName = $firstName;
return $this;
}
public function getLastName(): ?string
{
return $this->lastName;
}
public function setLastName(string $lastName): self
{
$this->lastName = $lastName;
return $this;
}
public function getPhone(): ?string
{
return $this->phone;
}
public function setPhone(string $phone): self
{
$this->phone = $phone;
return $this;
}
public function getBooking(): ?Booking
{
return $this->booking;
}
public function setBooking(?Booking $booking): self
{
$this->booking = $booking;
return $this;
}
public function getEmail(): ?string
{
return $this->email;
}
public function setEmail(?string $email): self
{
$this->email = $email;
return $this;
}
}
最后是我的 VisitorType 表单(有点简化):
class VisitorType extends AbstractType
{
public function buildForm(FormBuilderInterface $builder, array $options)
{
$builder
->add('firstName', TextType::class, [
'label' => 'entity.visitor.first-name',
])
->add('lastName', TextType::class, [
'label' => 'entity.visitor.last-name',
])
->add('phone', TextType::class, [
'label' => 'entity.visitor.phone-number',
'required' => false,
])
->add('email', TextType::class, [
'label' => 'entity.visitor.email',
'required' => false,
'constraints' => [
new Email()
]
])
;
}
public function configureOptions(OptionsResolver $resolver)
{
$resolver->setDefaults([
'data_class' => Visitor::class,
]);
}
}
我尝试向电子邮件和电话字段添加一个表达式约束,如下所示:
new Expression([
'expression' => 'this.getPhone() == null && this.getEmail() == null'
])
还尝试直接向实体添加约束,但对我来说似乎没有任何工作正常。
任何帮助将不胜感激。
更新 我没有指定这一点,但我的问题来自于我想从另一个将 VisitorType 添加为 CollectionType 的表单验证 VisitorType 表单。
【问题讨论】:
标签: forms constraints symfony4