如何在joomla 2.5前端發佈新創建的組件2.5

Question

我第一次創建了一個新組件。其後端功能都運行良好。現在我想在前臺展示。我創建了一個新的菜單項並選擇了這個新組件，但是在前端，當我點擊我的菜單Error 500與View not found [name, type, prefix]: tenders, html, tendersView。 我的網站的文件夾結構是：

joomla 
    |components 
    |com_tenders 
     |controller.php 
     |tenders.php 
     |models 
      |tenders.php 
     |views 
      |tenders 
       |view.html.php 
       |tmpl 
        |default.php 

com_tenders/Controller.php這樣

<?php 
defined('_JEXEC') or die('Restricted access'); 

jimport('joomla.application.component.controller'); 

class TendersController extends JController { 

     public function display() { 

     $input = JFactory::getApplication()->input; 
     $input->set('view', $input->getCmd('view', 'Tenders')); 
     parent::display(); 
     } 
} 
?> 

的意見/標段/ view.html.php

<?php 
defined('_JEXEC') or die('Restricted access'); 

jimport('joomla.application.component.view'); 

class TendersViewTenders extends JView { 

    function display($tpl = null) { 

    $model =& $this->getModel(); 
    $msg = $model->getMsg(); 
    $this->assignRef('msg',$msg); 

    parent::display($tpl); 

    } 
} 
?> 

型號/標段.php

<?php 
defined('_JEXEC') or die('Restricted access'); 
jimport('joomla.application.component.modelitem'); 

class TendersModelTenders extends JModelItem { 

    public function getMsg() { 
      $db =& JFactory::getDBO(); 
      $query = "SELECT * FROM #__tenders"; 
      $db->setQuery($query); 
      $msg = $db->loadResult(); 

      return $msg; 
    } 
} 
?> 

請讓我知道是否需要任何澄清。

Answer 1

在的意見/標段/ view.html.php重命名類從TendersViewTender到TendersViewTenders

你需要有tenders.php文件過你com_tenders文件夾。這應該看起來像這樣：

defined('_JEXEC') or die('Restricted access'); 
require_once (JPATH_COMPONENT.DS.'controller.php'); 

$controller = new TendersController(); 

$controller->execute(JRequest::getCmd('task')); 
$controller->redirect();