我正在尝试使用Codeigniter中的Mysql数据库开发4级动态菜单。主菜单和子菜单正常显示,但第三级和第四级没有显示。谁能帮我吗?
<?php
foreach($main_menu as $mmenu): ?>
<?php if($mmenu->main_id != $mmenu->item_main_id): ?>
<li>
<a href='<?php echo base_url(); ?><?php echo $mmenu->m_name; ?>'><?php echo $mmenu->m_name; ?></a>
</li>
<?php else: ?>
<li class='has-sub'><a href='#'><?php echo $mmenu->m_name; ?></a>
<ul>
<?php foreach($sub_menu as $smenu): ?>
<?php if($mmenu->main_id == $smenu->m_id): ?>
<li class='has-sub'>
<a href='<?php echo base_url(); ?><?php echo $smenu->m_item_name; ?>'><?php echo $smenu->m_item_name; ?></a>
<ul>
<?php foreach($sub_submenu as $ssmenu): ?>
<?php if($smenu->sub_id != $ssmenu->m_item_id): ?>
<?php else: ?>
<li class='has-sub'><a href='<?php echo base_url(); ?><?php echo $ssmenu->m_item_sub_name; ?>'><?php echo $ssmenu->m_item_sub_name; ?></a>
<ul>
<?php foreach($sub_menu_second as $sssmenu): ?>
<?php if($ssmenu->submenu_id == $sssmenu->m_item_sub_id): ?>
<li class="has-sub">
<a class="pl-5" href=''><?php echo $sssmenu->m_item_sub_item_name; ?></a>
</li>
<?php elseif(empty($sssmenu->m_item_sub_id)): ?>
<li>
<a class="pl-5" href=''><?php echo $sssmenu->m_item_sub_item_name; ?></a>
</li>
<?php endif; ?>
<?php endforeach; ?>
</ul>
</li>
<?php endif; ?>
<?php endforeach; ?>
</ul>
</li>
<?php else; ?>
<li><a href='<?php echo base_url(); ?><?php echo $smenu->m_item_name; ?>'><?php echo $smenu->m_item_name; ?></a></li>
<?php endif; ?>
<?php endforeach; ?>
</ul>
</li>
<?php endif ?>
<?php endforeach; ?>