我需要根据网站的父页面(或部分)拉一个子菜单。我有三个部分(我们的客栈、餐厅、活动),每个部分都需要一个单独的子菜单,我的客户可以通过WordPress的原生菜单部分进行编辑。
这是我的header.php
拉动菜单。
<?php
if (is_front_page()) {
wp_nav_menu( array(
\'theme_location\' => \'our-inn-menu\',
\'menu_id\' => \'sub-menu\',
\'container_class\' => \'sub-menu\',
) );
} elseif ($post->post_parent == 4) {
//Parent is Our Inn
wp_nav_menu( array(
\'theme_location\' => \'our-inn-menu\',
\'menu_id\' => \'sub-menu\',
\'container_class\' => \'sub-menu\',
) );
} elseif ($post->post_parent == 6) {
// Restaurant Page
wp_nav_menu( array(
\'theme_location\' => \'our-restaurant-menu\',
\'menu_id\' => \'sub-menu\',
\'container_class\' => \'sub-menu\',
) );
} elseif ($post->post_parent == 8) {
// Activities Page
wp_nav_menu( array(
\'theme_location\' => \'activities-menu\',
\'menu_id\' => \'sub-menu\',
\'container_class\' => \'sub-menu\',
) );
} else {
//Everything Else
wp_nav_menu( array(
\'theme_location\' => \'our-inn-menu\',
\'menu_id\' => \'sub-menu\',
\'container_class\' => \'sub-menu\',
) );
}
问题是,当我在后端分配每个菜单时,它只显示else
条款,无其他规定。当我移除else
子句,它显示一个空白div
.