如何删除“<span class =”caret“></span>”如果未在父类别下添加子类别

Question

我必须隐藏下拉“插入符号”。如果不在codeigniter中添加子类别。

这里是我的代码： ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

<ul class="nav navbar-nav"> 
<?php 
$this->load->module('store_categories'); 
foreach ($parent_categories as $key => $value) { 
    $parent_cat_id = $key; 
    $parent_cat_title = $value; 
?> 
    <li class="dropdown"> 
     <a href="#" class="dropdown-toggle" data-toggle="dropdown" role="button" aria-haspopup="true" aria-expanded="true"><?= $parent_cat_title ?> 
     <span class="caret"></span> 
     </a> 
     <ul class="dropdown-menu"> 
      <?php 
      $query = $this->store_categories->get_where_custom('parent_cat_id', $parent_cat_id); 
      foreach ($query->result() as $row) { 
      echo '<li><a href="#">'.$row->cat_title.'</a></li>'; 
      } 
      ?> 
     </ul> 
    </li> 
    <?php 
    } 
    ?> 
</ul> 

Answer 1

写一个函数生成查询内容，然后检查并使用它的值。

<?php 
    function getQueryContent($parent_cat_id, $that) { 
     $queryContent = ""; 
     $query = $that->store_categories->get_where_custom('parent_cat_id', $parent_cat_id); 
     foreach ($query->result() as $row) { 
      $queryContent .= '<li><a href="#">'.$row->cat_title.'</a></li>'; 
     } 
     return $queryContent; 
    } 
?> 

<ul class="nav navbar-nav"> 
<?php 
$this->load->module('store_categories'); 
foreach ($parent_categories as $key => $value) { 
    $parent_cat_id = $key; 
    $parent_cat_title = $value; 
    $queryContent = getQueryContent($parent_cat_id, $this); 
?> 
    <li class="dropdown"> 
     <a href="#" class="dropdown-toggle" data-toggle="dropdown" role="button" aria-haspopup="true" aria-expanded="true"><?= $parent_cat_title ?> 
     <?php 
     if ($queryContent) { 
     ?> 
     <span class="caret"></span> 
     <?php 
     } 
     ?> 
     </a> 
     <ul class="dropdown-menu"> 
      <?php 
      echo $queryContent; 
      ?> 
     </ul> 
    </li> 
    <?php 
    } 
    ?> 
</ul>