访问子子类别,Shopware,Smarty

时间:2018-01-19 10:25:22

标签: smarty shopware

试图通过smarty解决Shopware中子子类别访问的问题,但根本不确定如何做到这一点。也许有人知道。

<!-- THIS WORKS -->

{if $Category.sub}
  {foreach $Category.sub as $sub}
    <li>
      <a href="#">
        <span itemprop="name">{$sub.name}</span>
      </a>
    </li>
  {/foreach}
{/if}

<!-- THIS DOESN'T WORK -->
{if $Category.sub.sub}
  {foreach $Category.sub.sub as $subsub}
    <li>
      <a href="#">
        <span itemprop="name">{$subsub.name}</span>
      </a>
    </li>
  {/foreach}
{/if}

1 个答案:

答案 0 :(得分:0)

请查看以上示例:

{foreach $Category.sub as $categoryItem}
    {foreach $categoryItem.sub as $subsub}
        <li>
            <a href="#">
                <span itemprop="name">{$subsub.name}</span>
            </a>
        </li>
    {/foreach}
{/foreach}

高级示例:

{foreach $Category.sub as $number => $categoryItem}
    {if $categoryItem.childrenCount}
        <ul id="submenu-{$number}" {if $categoryItem.flag}class="is--active"{/if}>
            {if $categoryItem.cmsHeadline}
                <li class="menu--list-item item--level-1">
                    <span class="menu--list-item-title">{$categoryItem.cmsHeadline}</span>
                </li>
            {/if}
            {foreach $categoryItem.sub as $subcategory}
                {if $subcategory.hideTop}
                    {continue}
                {/if}

                {$categoryLink = $subcategory.link}
                {if $subcategory.external}
                    {$categoryLink = $subcategory.external}
                {/if}

                <li class="menu--list-item item--level-1{if $subcategory.flag} item--is-active{/if}">
                    <a href="{$categoryLink|escapeHtml}" class="menu--list-item-link" title="{$subcategory.name|escape}">{$subcategory.name}</a>
                </li>
            {/foreach}
        </ul>
    {/if}
{/foreach}