Question

我如何坚持为以下代码编写测试。我想模拟$ userModel，但是如何将其添加到测试中呢？

class PC_Validate_UserEmailDoesNotExist extends Zend_Validate_Abstract
{
    public function isValid($email, $context = NULL)
    {
        $userModel = new Application_Model_User();
        $user = $userModel->findByEmailReseller($email, $context['reseller']);

        if ($user == NULL) {
            return TRUE;
        } else {
            return FALSE;
        }
    }
}

更新：解决方案

我确实将我的类更改为以下内容以使其可测试，现在它使用依赖注入。有关依赖注入的更多信息，请找出here

我现在打电话给这个班：

new PC_Validate_UserEmailDoesNotExist(new Application_Model_User()

重构的课程

class PC_Validate_UserEmailDoesNotExist extends Zend_Validate_Abstract
{
    protected $_userModel;

    public function  __construct($model)
    {
        $this->_userModel = $model;
    }

    public function isValid($email, $context = NULL)
    {
        if ($this->_userModel->findByEmailReseller($email, $context['reseller']) == NULL) {
            return TRUE;
        } else {
            return FALSE;
        }
    }
}

单元测试

class PC_Validate_UserEmailDoesNotExistTest extends BaseControllerTestCase
{
    protected $_userModelMock;

    public function setUp()
    {
        parent::setUp();
        $this->_userModelMock = $this->getMock('Application_Model_User', array('findByEmailReseller'));
    }

    public function testIsValid()
    {
        $this->_userModelMock->expects($this->once())
                        ->method('findByEmailReseller')
                        ->will($this->returnValue(NULL));

        $validate = new PC_Validate_UserEmailDoesNotExist($this->_userModelMock);
        $this->assertTrue(
                $validate->isValid('jef@test.com', NULL),
                'The email shouldn\'t exist'
        );
    }

    public function testIsNotValid()
    {
        $userStub = new \Entities\User();

        $this->_userModelMock->expects($this->once())
                        ->method('findByEmailReseller')
                        ->will($this->returnValue($userStub));

        $validate = new PC_Validate_UserEmailDoesNotExist($this->_userModelMock);
        $this->assertFalse(
                $validate->isValid('jef@test.com', NULL),
                'The email should exist'
        );
    }
}

Answer 1

简短的回答：你不能，因为你将依赖项硬编码到方法中。

有三种解决方法：

1）使用的类名可配置，因此您可以执行以下操作：

$className = $this->userModelClassName;
$userModel = new $className();

或2）在方法签名中添加第三个参数，允许传入依赖项

public function isValid($email, $context = NULL, $userModel = NULL)
{
    if($userModel === NULL)
    {
        $userModel = new Application_Model_User();
    }
    // ...
}

或3）按照

中的描述使用set_new_overload()

注意： the Test-Helper extension is superseded https://github.com/krakjoe/uopz

PHPUnit如何模拟新创建的模型

1 个答案: