我是Joomla的新手,尤其是组件开发。无论如何,这是我的问题:
站点\视图\普通库\TMPL\默认.xml
<?xml version="1.0" encoding="utf-8"?>
<metadata>
<layout title="COM_PLAINGALLERY_PLAINGALLERY_VIEW_DEFAULT_TITLE">
<message>
<![CDATA[COM_PLAINGALLERY_PLAINGALLERY_VIEW_DEFAULT_DESC]]>
</message>
</layout>
<fields name="request"
addfieldpath="/administrator/components/com_plaingallery/models/fields">
<fieldset name="request">
<field name="galleryFolder" type="folderlist" default="" recursive="true"
label="Select a folder" directory="images" filter="" exclude="" width="300"
hide_none="true" hide_default="true" stripext="" />
</fieldset>
</fields>
</metadata>
站点\视图\普通库\视图.html.php
<?php
// No direct access to this file
defined('_JEXEC') or die('Restricted access');
// import Joomla view library
jimport('joomla.application.component.view');
/**
* HTML View class for the PlainGallery Component
*/
class PlainGalleryViewPlainGallery extends JViewLegacy
{
// Overwriting JView display method
function display($tpl = null)
{
// Assign data to the view
$this->msg = 'I am new to Joomla';
// Display the view
parent::display($tpl);
}
}
我的问题是:如何从用户在菜单配置中提供的字段[name="galleryFolder"]访问值?
感谢您的帮助!我真的很感激。
此参数位于菜单项的查询变量中。
例如,您可以尝试以下方法:
$app = JFactory::getApplication();
/* Default Page fallback*/
$active = $app->getMenu()->getActive();
if (NULL == $active) {
$active = $app->getMenu()->getDefault();
}
if ( isset($active->query['galleryFolder']) ) {
$galleryFolder = $active->query['galleryFolder'];
}