代码
class AclRowLevelsController extends AppController {
public $components = array(
// Don't use same name as Model
'_AclRowLevel' => array('className' => 'AclRowLevel')
);
public function view() {
$this->_AclRowLevel->checkUser();
...
}
}
class AclRowLevelComponent extends Component {
public function initialize(Controller $controller) {
$this->controller = $controller;
$this->AclRowLevel = ClassRegistry::init('AclRowLevel');
}
public function checkUser($permission, $model) {
$row = $this->AclRowLevel->find('first', array(
'conditions' => array(
'model' => $model['model'],
'model_id' => $model['model_id'],
'user_id' => $this->controller->Auth->user('id')
)
));
}
}
class AclRowLevelsControllerTest extends ControllerTestCase {
public function testViewAccessAsManager() {
$AclRowLevels = $this->generate('AclRowLevels', array(
'components' => array(
'Auth' => array(
'user'
),
'Session',
)
));
$AclRowLevels->Auth
->staticExpects($this->any())
->method('user')
->with('id')
->will($this->returnValue(1));
$this->testAction('/acl_row_levels/view/Task/1');
}
问题
AclRowLevel组件中的查询需要Auth用户ID。我想模拟user_id值' 1'用于单元测试。 模拟的Auth方法'用户'在我的测试中不适用于来自组件的调用。因此该查询中的用户标识值为null。
应该怎么做?
答案 0 :(得分:0)
执行debug($AclRowLevels->Auth);
检查是否真的被嘲笑了。它应该是一个模拟对象。如果不是出于某种原因尝试:
$AclRowLevels->Auth = $this->getMock(/*...*/);
checkUser()中的代码应该顺便进入模型。我也怀疑这必须是一个组成部分。这似乎是用于授权,所以为什么不making it a proper authorization adapter?
答案 1 :(得分:0)
这就是我想要的:
$AclRowLevels->Auth
->staticExpects($this->any())
->method('user')
->will($this->returnCallback(
function($arg) {
if ($arg === 'id') {
return 1;
}
return null;
}
));