如果条件匹配则应用约束

Question

我有一个 Symfony 表单。在这个表格中，我有两个字段：

"House number" 和 "Po Box"。它们的定义如下：

$builder->add('houseNumber', TextType::class, [
    'label' => 'Huisnummer',
    'attr' => [
        'maxlength' => 8,
    ],
    'constraints' => [
        new NotBlank([
            'groups' => $options['constraint_groups']
        ]),
        new Regex([
            'pattern' => '/^[0-9]+$/',
            'message' => 'Vul alleen het huisnummer in cijfers in.'
        ]),
        new Length([
            'groups' => $options['constraint_groups'],
            'max' => 8
        ])
    ]
])->add('poBox', TextType::class, [
    'label' => 'Postbus',
    'attr' => [
        'maxlength' => 10,
    ],
    'constraints' => [
        new Length([
            'groups' => $options['constraint_groups'],
            'max' => 10
        ])
    ]
]);

有没有办法让我在有 PoBox 的情况下不需要门牌号，反之亦然？

谢谢。

Answer 1

您可以创建自定义验证约束

namespace App\Validator\Constraints;

use Symfony\Component\Validator\Constraint;

/**
 * @Annotation
 * @Target({"CLASS", "ANNOTATION"})
 */
class HouseNumber extends Constraint
{
    public $message = 'Some message';

    public function getTargets()
    {
        return self::CLASS_CONSTRAINT;
    }
}

验证者

namespace App\Validator\Constraints;

use Symfony\Component\Form\Exception\UnexpectedTypeException;
use Symfony\Component\Validator\Constraint;
use Symfony\Component\Validator\ConstraintValidator;

class HouseNumberValidator extends ConstraintValidator
{
    public function validate($obj, Constraint $constraint)
    {
        if (!$constraint instanceof HouseNUmber) {
            throw new UnexpectedTypeException($constraint, __NAMESPACE__ . '\HouseNUmber');
        }

        $error = null;

        if (!$this->isValid($obj)) {
            $error = $constraint->message;
        }

        if (!empty($error)) {
            $this->context->buildViolation($error)
                ->atPath('houseNumber')
                ->addViolation();
        }
    }

    private function isValid($obj)
    {    
        if (!$obj instanceof SomeClass) {
            throw new UnexpectedTypeException($obj, SomeClass::class);
        }

        if (!empty($obj->getPoBox())) {
            return true;
        }

        return !empty($obj->getHouseNumber());
    }
}

然后在 class 中的 $houseNumber 字段中添加注释

use App\Validator\Constraints as AppAssert;

/**
 * @AppAssert\HouseNumber
 */
class SomeClass 
{
    private $houseNumber;
    ...
}

Answer 2

我认为实现您想要的效果的最佳方法是创建自定义 class 约束验证器 (https://symfony.com/doc/current/validation/custom_constraint.html#class-constraint-validator)。

我不知道你是否可以在表单上添加这个约束，或者你是否必须将你的表单与实体相关联。

如果条件匹配则应用约束

Apply constraint if conditions match

php

formbuilder

symfony