问当数据无效时，Symfony3单元测试表单总是有效的
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Stack Overflow用户

提问于 2017-09-30 04:20:05

回答 1查看 1.1K关注 0票数 0

我使用Symfony3和表单。我有一个formType，我想测试它。我遵循这篇文章：testing.html

我的formType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()),

            ));
    }
}

名字必须有三个以上的字符和年龄必须大于18。我想测试与phpunit，与无效的数据提交。年龄i submut是无效的，因为它是小于18，而名称是无效的，因为它包含一个字符。我的测试课程如下：

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());

    }
}

当我想用要提交的无效数据运行测试时，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？

php

forms

symfony

unit-testing

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回答 1

Stack Overflow用户

回答已采纳

发布于 2017-09-30 04:23:30

表单总是有效的，因为您的new ConstraintViolationList()是空的。

我可以建议您用以下方法替换UserTypeTest::getExtensions()方法：

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);
}