CSS获取没有课程的最后一个孩子

对于您来说，这是一个棘手的挑战，CSS选择器获取没有类的:last-child.

Here is a tricky challenge for you guys, CSS selector to get the :last-child that doesn't have a class.

到目前为止我尝试过的事情:

What I have tried so far:

.nav-item:not(.nav-item--mobile):last-child {...}
.nav-item:last-child:not(.nav-item--mobile) {...}

我有类似的查询选择器，它们可以做一些有趣的事情，所以我宁愿尝试通过CSS来做到这一点.流动物品的数量可以变化.

I have similar query selectors that do some fun stuff, so I'd rather try and do this via CSS. The mobile items can be variable in quantity.

// Get the first child in the list when there are n or more, and
// they are not mobile. Yes, this actually works.
.nav-item:first-child:nth-last-child(n+7):not(.nav-item--mobile)

在所有情况下，以下内容将给我最后一个孩子，但我希望最后一个不是仅移动孩子的孩子.

The following will give me the last child in all cases, but I want the last child that isn't a mobile only child.

.navigation-item--top:last-child

目标

generic generic generic generic generic mobile mobile
                                   ^
                             target this one

HTML

<ul class="nav-items-list">
    <li class="nav-item"></li>
    <li class="nav-item"></li>
    <li class="nav-item"></li>
    <li class="nav-item"></li>
    <li class="nav-item nav-item--mobile"></li>
    <li class="nav-item nav-item--mobile"></li>
</ul>

是的，我可以找出生成的导航中哪个是正确的，或者可以使用JS找到它.