CakePHP控制器的真实测试？

我正在使用CakePHP（刚刚发布的1.2.4）编写一个新的应用程序，使用SimpleTest 1.0.1。我已经阅读了的相关章节，搜索了，并阅读了Mark Story关于控制器测试的帖子（和）

不幸的是，这些都没有涉及到非平凡控制器的实际测试。许多应用程序将站点的各个区域置于登录之后，但我不知道如何测试以下简单场景：

• 来宾访问受保护页面重定向
• 有效凭据是否设置了预期的会话变量
• 无效凭据是否重新显示登录页面并显示错误消息

下面的控制器和测试不像我想的那样工作。两个断言都失败了，我还收到一个PHP错误：

失败 [NULL]在[…/app/tests/cases/controllers/users\u controller.test.php第79行不应为NULL] …/app/tests/cases/controllers/users\u controller.test.php->users controllertest->testLogin

失败 相等期望失败，因为在[…/app/tests/cases/controllers/users_controller.test.php第80行]处[NULL]与[Integer:1]不匹配 …/app/tests/cases/controllers/users\u controller.test.php->users controllertest->testLogin

错误 […/cake/libs/controller/components/auth.PHP第266行中出现意外PHP错误[Undefined index:action]严重性[E_NOTICE] …/app/tests/cases/controllers/users\u controller.test.php->users controllertest->testLogin

以下是控制器（baked plus Mark Story的“艰难之路”测试方法）：

以下是测试：

/* SVN FILE: $Id$ */
/* UsersController Test cases generated on: 2009-08-05 17:08:03 : 1249507923*/
App::import('Controller', 'Users');

class TestUsers extends UsersController
{
  var $autoRender = false;
  var $redirectUrl;
  var $redirectStatus;
  var $renderedAction;
  var $renderedLayout;
  var $renderedFile;
  var $stopped;

  function redirect($url, $status = null, $exit = true)
  {
    $this->redirectUrl = $url;
    $this->redirectStatus = $status;
  }

  function render($action = null, $layout = null, $file = null)
  {
    $this->renderedAction = $action;
    $this->renderedLayout = (is_null($layout) ? $this->layout : $layout);
    $this->renderedFile = $file;
  }

  function _stop($status = 0)
  {
    $this->stopped = $status;
  }
}

class UsersControllerTest extends CakeTestCase
{
  var $fixtures = array('user');
  var $Users = null;

  function startTest()
  {
    $this->Users = new TestUsers();
    $this->Users->constructClasses();
    $this->Users->Component->initialize($this->Users);
  }

  function prepareForAction()
  {
    $this->Users->beforeFilter();
    $this->Users->Component->startup($this->Users);
  }

  function endTest()
  {
    $this->Users->Session->destroy();
    unset($this->Users);
    ClassRegistry::flush();
  }

  //-----------------------------------------------------------------------

  function testUsersControllerInstance()
  {
    $this->assertTrue(is_a($this->Users, 'UsersController'));
  }

  function testLogin()
  {
    $this->Users->data = array(
      'User' => array(
        'username' => 'admin',
        'password' => 'admin'
      )
    );

    $this->prepareForAction();
    $this->Users->login();

    $this->assertNotNull($this->Users->redirectUrl);
    $this->assertEqual($this->Users->Session->read('Auth.User.id'), 1);
  }
}

---

您所做的测试并不是真正测试您的用户Contoller，而是真正测试AuthComponent。如果你想这样做，你需要确保你的TestUsersController的设置与你的应用程序相同。对于testLogin，您需要设置控制器的操作和url：

function testLogin()
{
 $this->Users->data = array(
              'User' => array(
                    'username' => 'admin',
                    'password' => 'admin'
                  )
            );

 $this->Users->params['url']['url'] = '/users/login';
 $this->Users->params['action'] = 'login';
 $this->prepareForAction();
 $this->Users->login();

 $this->assertNotNull($this->Users->redirectUrl);
 $this->assertEqual($this->Users->Session->read('Auth.User.id'), 1);
}

或者，我建议再看一眼，并使用这些方法为控制器代码编写测试，并模拟auth组件

function testLogin()
{
 $this->Users->data = array(
              'User' => array(
                    'username' => 'admin',
                    'password' => 'admin'
                  )
            );

 $this->Users->params['url']['url'] = '/users/login';
 $this->Users->params['action'] = 'login';
 $this->prepareForAction();
 $this->Users->login();

 $this->assertNotNull($this->Users->redirectUrl);
 $this->assertEqual($this->Users->Session->read('Auth.User.id'), 1);
}