Javascript 从li标签打开ul标签单击非全部-Opencart

有人能帮我使用corect js的手机菜单吗

所以我有这个代码：

<ul class="mob-menu-block">
           {% if categories %}
                            {% for category in categories %} 
                            {% if category.children %}
                             <li class="menu-link">
                                <a href="{{ category['href'] }}" data-toggle="dropdown" class="dropdown-toggle"><img src="{{ category['thumb'] }}" class="imgmenucat" loading="lazy" alt="{{ category['name'] }}" title="{{ category['name'] }}" />{{ category['name'] }}<i class="fas fa-angle-right"></i></a>
                             {% for children in category.children|batch(category.children|length / category.column|round(1, 'ceil')) %}
                                <ul class="dropdown-menu sub-block">
                                   <li><a href="{{ category['href'] }}" class="see-all">{{ text_all }} {{ category['name'] }}</a></li>
                                   <li role="separator" class="divider"></li>
                                  {% for child in children %}
                                     {% if child.children %}
                                   <li class="submenu-link">
                                      <a href="{{ child['href'] }}" data-toggle="dropdown" class="dropdown-toggle">{{ child['name'] }}<i class="fas fa-angle-right"></i></a>
                                      <ul class="dropdown-menu sub-block">
                                         <li><a href="{{ child['href'] }}" class="see-all">{{ text_all }} {{ child['name'] }}</a></li>
                                         <li role="separator" class="divider"></li>
                                              {% for children in child.children %}
                                         <li><a href="{{ child['href'] }}">{{ child['name'] }}</a></li>
                                          {% endfor %}    
                                      </ul>
                                   </li>
                                        {% else %}
                                   <li><a href="{{ child['href'] }}">{{ child['name'] }}</a></li>
                                {% endif %}
                                    {% endfor %}    
                                </ul>
                               {% endfor %}         
                             </li>
                            {% else %}
                             <li class="menu-link"><a href="{{ category['href'] }}"><img src="{{ category['thumb'] }}" class="imgmenucat" loading="lazy" alt="{{ category['name'] }}" title="{{ category['name'] }}"/>{{ category['name'] }}</a></li>
                             {% endif %}
                              {% endfor %}                                        
                {% endif %}
             </ul>

我不知道这是否正确。。。因为当我点击任何子类别时，它会打开所有带有子块的div，包括内部div。。。它必须只打开打开的div

打印屏幕：

有人能帮我吗

$(document).ready(function() {
    $('#menu').click(function() {
            $('.menu-block').slideToggle("fast");
    });
      $('#menu-mob').click(function() {
            $('.mob-menu-block').slideToggle("fast");
    });
    $('.dropdown-toggle').click(function() {
            $('.sub-block').slideToggle("fast");
    });
});