php – joomla 1.7访问：受保护的菜单参数

我想访问joomla 1.7中菜单的menu_image参数

Normally I just use:
$currentMenuItem = JSite::getMenu()->getActive();

它给了我所有参数如下,但是params对象中的数据是：protected所以我无法访问$currentMenuItem-> params-> data：protected-> menu_image因为我无法使用：在一个对象中.

Any1知道怎么做呢？

[params] => JRegistry Object
        (
            [data:protected] => stdClass Object
                (
                    [show_title] => 
                    [link_titles] => 
                    [show_intro] => 
                    [show_category] => 
                    [link_category] => 
                    [show_parent_category] => 
                    [link_parent_category] => 
                    [show_author] => 
                    [link_author] => 
                    [show_create_date] => 
                    [show_modify_date] => 
                    [show_publish_date] => 
                    [show_item_navigation] => 
                    [show_vote] => 
                    [show_icons] => 
                    [show_print_icon] => 
                    [show_email_icon] => 
                    [show_hits] => 
                    [show_noauth] => 
                    [menu-anchor_title] => 
                    [menu-anchor_css] => 
                    [menu_image] => images/joomla_black.gif
                    [menu_text] => 1
                    [page_title] => 
                    [show_page_heading] => 0
                    [page_heading] => 
                    [pageclass_sfx] => 
                    [menu-meta_description] => 
                    [menu-meta_keywords] => 
                    [robots] => 
                    [secure] => 0
                )

        )

解决方法:

正如@tereško指出的那样：你扩展了类并创建了一个getter.好吧obv joomla allready有一个getter：所以这里去(例如获取受保护的menu_image)

$currentMenuItem = JSite::getMenu()->getActive();
$currentMenuItem->params->get('menu_image', 'images/no-image.jpg');

标签：php,oop,joomla,joomla1-7
来源： https://codeday.me/bug/20190518/1126596.html