可捕捉的致命错误 - Joomla Compoent - FOF

Question

我使用框架在框架上新建Joomla组件。管理员部分工作正常。在站点部分显示以下错误。如何解决此问题错误。

Catchable fatal error: Argument 1 passed to FOFTable::setInput() must be an instance of FOFInput, instance of F0FInput given, called in /var/www/testjoomla/libraries/f0f/table/table.php on line 434 and defined in /var/www/testjoomla/libraries/fof/table/table.php on line 3236

在我的调度代码

：

include_once JPATH_LIBRARIES.'/fof/include.php'; 
class GulfJobDispatcher extends FOFDispatcher 
{ 
    public function onBeforeDispatch() { 
     $result = parent::onBeforeDispatch(); 

     if($result) { 
      // Load Akeeba Strapper 
      include_once JPATH_ROOT.'/media/akeeba_strapper/strapper.php'; 
      AkeebaStrapper::bootstrap(); 
      AkeebaStrapper::jQueryUI(); 
      AkeebaStrapper::addCSSfile('media://com_gulfjob/css/frontend.css'); 
     } 

     return $result; 
    } 
} 

Answer 1

变化F0F到FOF在您的控制器或其他地区