Symfony3 单元测试表单在无效数据时始终有效

Question

我使用 Symfony3 和表单。我有一个 formType，我想测试它。 我关注这篇文章： http://symfony.com/doc/current/form/unit_testing.html

我的表单类型如下：

namespace AppBundle\Form;

use Symfony\Component\Form\AbstractType;
use Symfony\Component\Form\Extension\Core\Type\IntegerType;
use Symfony\Component\Form\Extension\Core\Type\TextType;
use Symfony\Component\Form\FormBuilderInterface;
use Symfony\Component\Validator\Constraints\GreaterThanOrEqual;
use Symfony\Component\Validator\Constraints\Length;
use Symfony\Component\Validator\Constraints\NotBlank;
use Symfony\Component\Validator\Constraints\Required;

class UserType extends AbstractType
{
    public function buildForm(FormBuilderInterface $builder, array $aOption)
    {

        $builder->add('name', TextType::class, array(
            'constraints' => array(
                new NotBlank(),
                new Length(array('min' => 3)),
            ),
        ))
            ->add('age', IntegerType::class, array(
                'constraints' => array(
                    new GreaterThanOrEqual(array('value' => 18)),
                    new Required()),

            ));
    }
}

姓名必须超过 3 个字符，年龄必须超过 18 岁。 我想用 phpunit 进行测试，提交无效数据。我 submut 的年龄无效，因为它小于 18，名称无效，因为它包含 1 个字符。我的测试类如下：

namespace Tests\AppBundle\Controller;

use AppBundle\Form\UserType;
use Symfony\Component\Form\Test\TypeTestCase;
use Symfony\Component\Form\Extension\Validator\ValidatorExtension;
use Symfony\Component\Form\Form;
use Symfony\Component\Validator\ConstraintViolationList;
use Symfony\Component\Validator\Mapping\ClassMetadata;
use Symfony\Component\Validator\Validator\ValidatorInterface;

class UserTypeTest extends TypeTestCase
{
    private $validator;
    protected function getExtensions()
    {
        $this->validator = $this->createMock(ValidatorInterface::class);
        // use getMock() on PHPUnit 5.3 or below
        // $this->validator = $this->getMock(ValidatorInterface::class);
        $this->validator
            ->method('validate')
            ->will($this->returnValue(new ConstraintViolationList()));
        $this->validator
            ->method('getMetadataFor')
            ->will($this->returnValue(new ClassMetadata(Form::class)));
        return array(
            new ValidatorExtension($this->validator),
        );
    }
    public function testIndex()
    {
        $formData = array(
            'name' => 'a',
            'age'  => '4'
        );
        $formBuilder = $this->factory->createBuilder(UserType::class);
        $form = $formBuilder->getForm();
        $form->submit($formData);
        $this->assertTrue($form->isSynchronized());
        $this->assertFalse($form->isValid());

    }
}

当我想使用无效数据运行测试以提交时，我失败了：

./vendor/bin/phpunit tests/AppBundle/Controller/UserTypeTest.php  
Failed asserting that true is false.

为什么表单总是有效的？

Answer 1

您的表单始终有效，因为您的 new ConstraintViolationList() 为空。

我可以建议您将 UserTypeTest::getExtensions() 方法替换为以下内容：

public function getExtensions()
{
    $extensions = parent::getExtensions();
    $metadataFactory = new FakeMetadataFactory();
    $metadataFactory->addMetadata(new ClassMetadata(  Form::class));
    $validator = $this->createValidator($metadataFactory);

    $extensions[] = new CoreExtension();
    $extensions[] = new ValidatorExtension($validator);

    return $extensions;
}


protected function createValidator(MetadataFactoryInterface $metadataFactory, array $objectInitializers = array())
{
    $translator = new IdentityTranslator();
    $translator->setLocale('en');
    $contextFactory = new ExecutionContextFactory($translator);
    $validatorFactory = new ConstraintValidatorFactory();
    return new RecursiveValidator($contextFactory, $metadataFactory, $validatorFactory, $objectInitializers);
}

有了这个 ValidatorExtension，你不应该为空 ConstraintViolationList