Mysql的動態下拉菜單結果集

Question

我想從下面的結果集中創建一個動態菜單。

HTML結構應該如下，

<ul class="menu"> 

<li>Menu 1</li> 
    <ul class='submenu'> 
    <li>Submenu1</li> 
    <li>Submenu2</li> 
    </ul>  
<li>Menu2</li> 
    <ul class='submenu'> 
    <li>Submenu1</li> 
    <li>Submenu2</li> 
    </ul> 
<li>Menu 3</li> 
</ul> 

嘗試作爲以下，

<?php $cat = 0;?> 
<?php foreach($this->submenus as $submenu): ?> 

     <?php if($cat!= $submenu->category_id): ?> 


         <li><?php echo $submenu->category_name ?></li> 

         <?php echo (!empty($submenu->subcategory_name))?'<ul>':'';  ?> 

         <?php $flag = $submenu->category_id; ?> 

     <?php endif;?> 

         <li class='sub'><?php echo $submenu->subcategory_name ?></li> 

     <?php if($cat!= $flag && $cat > 0): ?> 
       </ul> 
     <?php endif;?>           

       <?php $cat = $submenu->category_id;?> 


<?php endforeach; ?> 

它未能當接近關閉。

請幫忙。

謝謝

Answer 1

我已經重新安排你的PHP代碼了一下，請嘗試以下方法：

<?php $cat = 0;?> 
<?php $submenu_flag = 0;?> 

<?php foreach($this->submenus as $submenu): ?> 

     <?php if($cat!= $submenu->category_id): ?> 


       <?php if($submenu_flag > 0): ?> 
         </ul> 
       <?php $submenu_flag = 0;?> 
       <?php endif;?>           

         <li><?php echo $submenu->category_name ?></li> 


       <?php if(!empty($submenu->subcategory_name)): ?> 
         <ul> 
          <?php $submenu_flag = 1; ?> 
       <?php endif;?>           


     <?php endif;?> 


      <?php echo (!empty($submenu->subcategory_name))?'<li class='sub'><?php echo $submenu->subcategory_name ?></li>':'';  ?> 


      <?php $cat = $submenu->category_id;?> 


<?php endforeach; ?> 

Answer 2

您可以使用Accordion菜單執行此任務。

請參閱此鏈接http://chandreshmaheshwari.wordpress.com/2011/05/27/accordian-menu/

感謝。