PHPUnit模拟对象和静态方法

Question

我正在寻找测试以下静态方法的最佳方法(特别是使用Doctrine模型):

class Model_User extends Doctrine_Record
{
    public static function create($userData)
    {
        $newUser = new self();
        $newUser->fromArray($userData);
        $newUser->save();
    }
}

理想情况下,我会使用模拟对象来确保调用"fromArray"(使用提供的用户数据)和"save",但这是不可能的,因为该方法是静态的.

有什么建议？

Answer 1

PHPUnit的作者Sebastian Bergmann最近发表了一篇关于Stubbing和Mocking Static Methods的博客文章.使用PHPUnit 3.5和PHP 5.3以及后期静态绑定的一致使用,你可以做到

$class::staticExpects($this->any())
      ->method('helper')
      ->will($this->returnValue('bar'));

更新: staticExpects被弃用的PHPUnit 3.8的,将被更高版本的完全去除.

Answer 2

现在有AspectMock库来帮助解决这个问题:

https://github.com/Codeception/AspectMock

$this->assertEquals('users', UserModel::tableName());   
$userModel = test::double('UserModel', ['tableName' => 'my_users']);
$this->assertEquals('my_users', UserModel::tableName());
$userModel->verifyInvoked('tableName'); 

Answer 3

我将在单元测试命名空间中创建一个新类来扩展Model_User并测试它。这是一个例子：

原班：

class Model_User extends Doctrine_Record
{
    public static function create($userData)
    {
        $newUser = new self();
        $newUser->fromArray($userData);
        $newUser->save();
    }
}

在单元测试中调用的模拟类：

use \Model_User
class Mock_Model_User extends Model_User
{
    /** \PHPUnit\Framework\TestCase */
    public static $test;

    // This class inherits all the original classes functions.
    // However, you can override the methods and use the $test property
    // to perform some assertions.
}

在您的单元测试中：

use Module_User;
use PHPUnit\Framework\TestCase;

class Model_UserTest extends TestCase
{
    function testCanInitialize()
    {   
        $userDataFixture = []; // Made an assumption user data would be an array.
        $sut = new Mock_Model_User::create($userDataFixture); // calls the parent ::create method, so the real thing.

        $sut::test = $this; // This is just here to show possibilities.

        $this->assertInstanceOf(Model_User::class, $sut);
    }
}