未找到控制器时,通过默认控制器路由zend请求

时间:2010-04-27 01:28:21

标签: model-view-controller zend-framework bootstrapping

以下是我Bootstrap课程中定义的函数。我必须错过Zend进行路由和调度的方式。我想要完成的任务很简单:对于任何因任何原因无法发送的请求/foo/bar/*,请尝试/index/foo/bar/。我遇到的问题是当FooController存在时,我得到Action "foo" does not exist。基本上,isDispatchable总是错误的。

public function run() {
        $front = Zend_Controller_Front::getInstance();
        $request = $front->getRequest();
        $dispatcher = $front->getDispatcher();
        //$controller = $dispatcher->getControllerClass($request);
        if (!$dispatcher->isDispatchable($request)) {
            $route = new Zend_Controller_Router_Route(
                ':action/*',
                array('controller' => 'index')
            );
            $router = $front->getRouter();
            $router->addRoute('FallBack', $route);
        }
        $front->dispatch();
    }

2 个答案:

答案 0 :(得分:0)

如果我理解你的想法是正确的 你会尝试使用__call魔术方法吗? 然后使用$this->_redirect();进行默认操作,例如

此处有更多信息http://php.net/manual/en/language.oop5.overloading.php

<强>更新

如果您在第480行打开Zend / Controller / Action.php

public function __call($methodName, $args)
    {
        require_once 'Zend/Controller/Action/Exception.php';
        if ('Action' == substr($methodName, -6)) {
            $action = substr($methodName, 0, strlen($methodName) - 6);
            throw new Zend_Controller_Action_Exception(sprintf('Action "%s" does not exist and was not trapped in __call()', $action), 404);
        }

        throw new Zend_Controller_Action_Exception(sprintf('Method "%s" does not exist and was not trapped in __call()', $methodName), 500);
    }

我打算做的是扩展这个类并完全覆盖__call函数

classs My_Controller_Action extends Zend_Controller_Action{
   public function __call($methodName, $args)
        {
            ///// do your magic here ......redirection or logging the request or what ever 
        }
}

并确保您的控制器扩展新创建的类<​​/ p>

class FooController extends My_Controller_Action
{
   public function indexAction()
    {
        // action body
    }
}

因此,如果您调用不存在的操作__call的某些方式将会运行 这个想法只是关于不存在的行为 如果控制器不存在,它将无法工作

答案 1 :(得分:0)

所以这似乎有效,但不是最好的答案,因为它只是放弃了所有的参数。我可能会尝试很快在插件中转发/index/[original uri]

class Bootstrap extends Zend_Application_Bootstrap_Bootstrap {
  protected function _initRoute() {
    $front = Zend_Controller_Front::getInstance();
    $routes = array(
      'FallBack' => new Zend_Controller_Router_Route(
        ':controller/:action/*',
        array('controller' => 'index', 'action' => 'index')
      )
    );
    $router = $front->getRouter();
    $router->removeDefaultRoutes();
    $router->addRoutes($routes);
    $front->setRouter($router);
    return $router;
  }

  protected function _initPlugin() {
    $front = Zend_Controller_Front::getInstance();
    $front->registerPlugin(new My_Controller_Plugin_FallBack());
  }
    }

class My_Controller_Plugin_FallBack extends Zend_Controller_Plugin_Abstract {
  public function preDispatch(Zend_Controller_Request_Abstract $request) {
    $front = Zend_Controller_Front::getInstance();
    $dispatcher = $front->getDispatcher();
    $router = $front->getRouter();
    if (($router->getCurrentRouteName() == 'FallBack') &&
        !$dispatcher->isDispatchable($request)) {
      $request->setActionName($request->getControllerName());
      $request->setControllerName('index');
    }
  }
}
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