所以我在我的 drupal 7 站点的一个区域中有几个菜单块。我需要将这些菜单块中的每一个都包装在一个<section>
标签中,但不影响所有其他菜单块。我的想法是预处理该区域,检查这些块是否是菜单块,并且如您所见,尝试简单地将输出包装在一个部分标记中。有人可以告诉我我做错了什么吗?这个问题正在杀死我。
function mytheme_preprocess_region(&$vars){
//checks to see if we're in the correct region
if($vars['region'] == "footer-top"){
//loops through every block in our region
foreach($vars['elements'] as $key => $item){
$block_type = $item['#block']->module;
//if it's a menu block, wrap the output in section tag, this doesnt work
if($block_type == "menu_block"){
//$vars['elements']['menu_block_4']['#children'] = "<section>" . $item['#children'] . "</section>";
$vars['elements'][$key]['#children'] = "<section>" . $item['#children'] . "</section>";
}
}
}
}