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我是在 PHPUnit 中模拟对象的新手，无法让它工作。我正在构建当前的扩展SensioGeneratorBundle（用于 Symfony2）。我使用通过PEAR. 它在PHP 5.3.5上运行（因为 PEAR 安装在该版本中）。

我的剥离课程是：

控制器生成器.php

class ControllerGenerator extends Generator
{
    // ...

    public function generate(BundleInterface $bundle, $controller, array $actions = array())
    {
        // ...
    }
}

生成控制器命令.php

class GenerateControllerCommand extends ContainerAwareCommand
{
    private $generator;

    /**
     * @see Command
     */
    public function configure()
    {
        // ...
    }

    public function execute(InputInterface $input, OutputInterface $output)
    {
        // ...

        $generator = $this->generator;
        $generator->generate($bundle, $controller);

        // ...
    }

    protected function getGenerator()
    {
        if (null === $this->generator) {
            $this->generator = new ControllerGenerator($this->getContainer()->get('filesystem'), __DIR__.'/../Resources/skeleton/bundle');
        }

        return $this->generator;
    }

    public function setGenerator(ControllerGenerator $generator)
    {
        $this->generator = $generator;
    }
}

生成控制器命令测试.php

class GenerateControllerCommandTest extends GenerateCommandTest
{
    public function testNonInteractiveCommand()
    {
        $bundle = 'FooBarBundle';
        $controller = 'PostController';

        $input = array(
            'command' => 'generate:controller',
            '--bundle' => $bundle,
            '--controller' => $controller,
        );

        $application = $this->getApplication();
        $commandTester = $this->getCommandTester($input);
        $generator = $this->getGenerator();

        $generator
            ->expects($this->once())
            ->method('generate')
            ->with($this->getContainer()->get('kernel')->getBundle($bundle), $controller)
        ;

        $commandTester->execute($input, array('interactive' => false));
    }

    protected function getCommandTester($input = '')
    {
        return new CommandTester($this->getCommand($input));
    }

    protected function getCommand($input = '')
    {
        return $this->getApplication($input)->find('generate:controller');
    }

    protected function getApplication($input = '')
    {
        $application = new Application();

        $command = new GenerateControllerCommand();
        $command->setContainer($this->getContainer());
        $command->setHelperSet($this->getHelperSet($input));
        $command->setGenerator($this->getGenerator());

        $application->add($command);

        return $application;
    }

    protected function getGenerator()
    {
        // get a noop generator
        return $this
            ->getMockBuilder('Sensio\Bundle\GeneratorBundle\Generator\ControllerGenerator')
            ->disableOriginalConstructor()
            ->setMethods(array('generate'))
            ->getMock()
        ;
    }
}

当我运行 PHPUnit 时，我不断收到此错误：

 $ phpunit Tests\Command\GenerateControllerCommandTest

     PHPUnit 3.7.0 by Sebastian Bergmann.

     Configuration read from E:\Wouter\web\wamp\www\wjsnip\vendor\sensio\generator-bundle\Sensio\Bundle\GeneratorBundle\phpunit.xml.dist

     F

     Time: 2 seconds, Memory: 7.25Mb

     There was 1 failure:

     1) Sensio\Bundle\GeneratorBundle\Tests\Command\GenerateControllerCommandTest::testNonInteractiveCommand
     Expectation failed for method name is equal to <string:generate> when invoked 1 time(s).
     Method was expected to be called 1 times, actually called 0 times.

     E:\Wouter\web\wamp\bin\php\php5.3.5\PEAR\phpunit:46

     FAILURES!
     Tests: 1, Assertions: 7, Failures: 1.

为什么我会收到此错误？我想我generate在方法中调用了命令GenerateControllerCommand::execute？我做错了什么，可能是真的吗？或者这是 PHPunit 中的错误？

4

1 回答 1

2

简而言之

您生成两个不同的$generator对象。通话发生在一个人身上，而另一个人接听了expect它。

更长

您更改了行为protected function getGenerator()，但原始函数期望调用该函数会填充$this->generator.

您的测试无法正常工作，并且该函数期望始终获得相同的生成器，并且随着您的覆盖，该函数将返回两个不同的对象。

您将预期的调用设置为一个，并且调用发生在对象上。

只是看着：

    $generator = $this->getGenerator();

    $generator
        ->expects($this->once())
        ->method('generate')
        ->with($this->getContainer()->get('kernel')->getBundle($bundle), $controller)
    ;

    $commandTester->execute($input, array('interactive' => false));
}

该$generator变量不会放在任何地方的任何对象范围内，因此不会发生任何调用，因为每次调用都会$this->getGenerator()产生一个未存储在任何地方的新对象。

所以在

protected function getApplication() {
    //...
    $command->setGenerator($this->getGenerator());
    //...
}

您的对象与测试用例中的对象不同。

于 2012-09-27T01:29:45.137 回答