Joomla組件：視圖：從字段獲取值

Question

我是Joomla的新手，尤其是組件開發。 無論如何，這是我的問題：

site \\ views \\ plaingallery \\ tmpl \\ default.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>

site \\ views \\ plaingallery \\ view.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);
    }
}

我的問題是：如何從菜單配置中提供的用戶從字段[name =“ galleryFolder”]中訪問值？

謝謝你的幫助！ 我真的很感激。

Answer 1

此參數位於菜單項的查詢變量中。

您可以嘗試例如：

    $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'];
    }