小编Gui*_*que的帖子

我正在寻找在layout.phtml文件中显示导航菜单:

<?php echo $this->navigation('navigation')->menu(); ?>

因此,我将这些行添加到/module/Application/config/module.config.php,以便根据我的应用程序路由声明菜单的结构:

'navigation' => array(
        'default' => array(
            'place' => array(
                'label' => 'Places',
                'route' => '/place',
                'pages' => array(
                    'country' => array(
                        'label' => 'Countries',
                        'route' => '/place/country',
                    ),
                    'state' => array(
                        'label' => 'States',
                        'route' => '/place/state',
                    ),
                    'city' => array(
                        'label' => 'Cities',
                        'route' => '/place/city',
                    ),
                ),
            ),
        ),
     ),

然后我为Zend/Navigation实例编写了一个加载器:

'service_manager' => array(
        'factories' => array(
            'translator' => 'Zend\I18n\Translator\TranslatorServiceFactory',
            'navigation' => 'Zend\Navigation\Service\DefaultNavigationFactory',
        ),
    ),

但我反复得到这个例外:

致命错误:Zend\Mvc\Router\Exception\RuntimeException:在C:\ wamp\www\bsol\vendor\zendframework\zendframework\library\Zend\View\Helper\Navigation\AbstractHelper.php中找不到名称为""的路由第471行

我试图解决将此行添加到菜单声明中的问题:

             'pages' => array( …