Question

如何覆盖模块的Yii登录URL模块？

这是基本应用程序的主要配置：

return array(
            .........

    // application components
    'components'=>array(
        'user'=>array(              
            // enable cookie-based authentication
            'allowAutoLogin'=>true,
            'loginUrl' => '/site/login',
        ),
           ..........
      );

然后我有代理模块，我希望在这个模块中登录URL不同，登录方法也不同。

class AgentModule extends CWebModule {

    public function init() {
        // this method is called when the module is being created
        // you may place code here to customize the module or the application
        // import the module-level models and components
        $this->setImport(array(
            'agent.models.*',
            'agent.components.*',
        ));

        $this->defaultController = 'default';
        $this->layoutPath = Yii::getPathOfAlias('agent.views.layout');
        $this->components = array(
            'user' => array(
                'class' => 'AgentUserIdentity',
                'loginUrl' => '/agent/default/login',
            )
        );
    }
            .......

但我不知道为什么这不起作用。请帮忙......（T.T）

Answer 1

使用此代码



class AgentModule extends CWebModule {
    public $assetsUrl;
    public $defaultController = 'Login';

    public function init() {

        // this method is called when the module is being created
        $this->setComponents(array(
                'errorHandler' => array(
                        'errorAction' => 'admin/login/error'),
                'user' => array(
                        'class' => 'CWebUser',
                        'loginUrl' => Yii::app()->createUrl('admin/login'),
                )
                )
        );

        Yii::app()->user->setStateKeyPrefix('_admin');

        // import the module-level models and components
        $this->setImport(array(
                'admin.models.*',
                'admin.components.*',
                )
        );
    }

    public function beforeControllerAction($controller, $action) {

        if(parent::beforeControllerAction($controller, $action)) {
            // this method is called before any module controller 
            //action is performed
            $route = $controller->id . '/' . $action->id;


            $publicPages = array(
                    'login/login',
                    'login/error',
            );

            if (Yii::app()->user->name !== 'admin' && !in_array($route, 
                              $publicPages)) {
                Yii::app()->getModule('admin')->user->loginRequired();
            } else {
                return true;
            }
        }
        else
            return false;
    }
}

Answer 2

在您的问题的第二个代码块中，您为模块user定义了AgentModule组件。

如果我理解正确，您在视图或控制器中的哪个位置使用Yii::app()->user->loginUrl，对吧？要在AgentModule中定义网址，您应该在Yii::app()->user->controller->module->user上下文中使用AgentModule但。

让它与众不同的最简单方法是为模块和app定义某个登录网址，并使用此预定义网址。

在此处阅读有关模块组件的信息：http://www.yiiframework.com/wiki/27/how-to-access-a-component-of-a-module-from-within-the-module-itself/

Answer 3

class WebUser extends CWebUser {

    public $module = array();

    public function init() {
        parent::init();
        if(isset($this->module['loginUrl'])){
            if(!isset($this->module['moduleId'])){
                throw new Exception('moduleId Must defined');
            }else{
                $moduleId = Yii::app()->controller->module->id;
                if($moduleId == $this->module['moduleId']){
                    #$this->loginUrl = Yii::app()->request->redirect(Yii::app()->createUrl($this->module['loginUrl']));
                    $this->loginUrl = array($this->module['loginUrl']);
                }
            }
        }

    }

}

after that you need to do some changes under config file i.e 
return array(
    'components' => array(
        'user' => array(
            'allowAutoLogin' => true,
            'class' => 'WebUser',
            'module' => array(
                'loginUrl' => '/r_admin',
                'moduleId' => 'r_admin'
            )
        ),
    )
);

覆盖模块的Yii登录URL

3 个答案: