在 Laravel 中活跃的动态 li 类

Question

这是我的路线<?php/*|--------------------------------------------------------------------------| Web Routes|--------------------------------------------------------------------------|| Here is where you can register web routes for your application. These| routes are loaded by the RouteServiceProvider within a group which| contains the "web" middleware group. Now create something great!|*/Route::get('/', function () {    return view('welcome');});Route::get('/account', 'AccountController@index')->name('account.index')->middleware('auth');Route::get('/dashboard', 'DashboardController@index')->name('dashboard.show')->middleware('auth');在我的仪表板上<div class="main-menu-content">        <ul class="navigation navigation-main" id="main-menu-navigation" data-menu="menu-navigation">          <li {{ request()->route()->getName() === 'dashboard.index' ? ' class=active' : '' }}><a href="/dashboard"><i class="ft-home"></i><span class="menu-title" data-i18n="">Dashboard</span></a>          </li>          <li {{ request()->route()->getName() === 'account.index' ? ' class=active' : '' }}><a href="/account"><i class="ft-user"></i><span class="menu-title" data-i18n="">Account</span></a>          </li>        </ul>      </div>我在这里要做的是使li课程处于活动状态，但是我尝试了上面的代码，尽管我在仪表板上但没有活动课程，而且我还检查了我在帐户中时li仍然没有突出显示为活动的。有人可以告诉我有什么问题吗？

Answer 1

您没有命名帐户路由，因此您必须将帐户路由更改为：

Route::get('/account', 'AccountController@index')->name('account.index')->middleware('auth');

您确实将仪表板路由命名为dashboard.show，因此li仪表板的标签必须等于dashboard.show而不是dashboard.index

Answer 2

<li class="@if(Route::currentRouteName() == 'dashboard.show' active @endif"><a href="/dashboard"><i class="ft-home"></i><span class="menu-title" data-i18n="">Dashboard</span></a>