Laravel 4 基于路线Laravel的包含

我如何包含一个基于哪个路由类别的文件，该类别与带Laravel的刀片一起使用？我基本上是在视图模板的子导航部分尝试这样做：

app/views/layouts/default.blade.php

@if (Route::resource() == 'tasks')
    @include('navs.task')
@elseif (Route::resource() == 'projs')
    @include('navs.proj')
@elseif (Route::resource() == 'miscs')
    @include('navs.misc')
@else 
    @include('navs.info')
@endif

---

但这会引发错误，未定义的类常量“resource”

您正在查找Route:：currentRouteName

@if (Route::currentRouteName() == 'tasks')
    @include('navs.task')
@elseif (Route::currentRouteName() == 'proj')
    @include('navs.proj')
@elseif (Route::currentRouteName() == 'misc')
    @include('navs.misc')
@else 
    @include('navs.info')
@endif

---

迈克尔p：你的回答让我找到了正确的答案。Route:：currentRouteName返回的“task.index”或“proj.create”过于精细。Route:：current->getUri有效

@if (Route::current()->getUri() == 'tasks')
    @include('nav.task')
@elseif (Route::current()->getUri() == 'projs')
    @include('nav.proj')
@elseif (Route::current()->getUri() == 'miscs')
    @include('nav.misc')
@else 
    @include('nav.info')
@endif

---

很晚了，但我认为使用Request:：is是一种更简单的方法

@if (Route::is('tasks'))
   @include('navs.task')
@elseif (Route::is('projs'))
   @include('nav.proj')
@elseif (Route::is('miscs'))
    @include('nav.misc')
@else 
    @include('nav.info')
@endif

此外，这也是if else的许多使用开关案例