如何在 laravel 中定义路由组名称

Question

有什么方法可以在 laravel 中定义路由组的名称吗？

我试图通过这个来完成的是知道当前请求属于哪个组，这样我就可以通过当前路由操作激活主菜单和子菜单：

代码：

Route::group(['prefix'=>'accounts','as'=>'account.'], function(){
    Route::get('/', 'AccountController@index')->name('index');
    Route::get('connect', 'AccountController@connect')->name('connect');
});

Route::group(['prefix'=>'quotes','as'=>'quote.'], function(){
    Route::get('/', 'QuoteController@index')->name('index');
    Route::get('connect', 'QuoteController@create')->name('create');
});

导航 HTML 代码

<ul>
    <li> // Add class 'active' when any route is open from account route group
        <a href="{{route('account.index')}}">Accounts</a>
        <ul>
            <li> // Add class 'active' when connect sub menu is clicked
                <a href="{{route('account.connect')}}">Connect Account</a>
            </li>
        </ul>
    </li>
    <li> // Add class 'active' when any route is open from quote route group
        <a href="{{route('quote.index')}}">Quotes</a>
        <ul>
            <li> // Add class 'active' when create sub menu is clicked
                <a href="{{route('quote.create')}}">Create Quote</a>
            </li>
        </ul>
    </li>
</ul>

现在我想要的是调用一个函数或其他东西，它会给我当前路由的组​​名。

例子：

1. 如果我在索引或创建引用页面getCurrentRouteGroup() 应该返回quote
2. 如果我在索引或帐户连接页面上getCurrentRouteGroup() 应该返回account

Answer 1

这应该可行：

Route::group(['prefix'=>'accounts','as'=>'account.'], function(){
    Route::get('/', ['as' => 'index', 'uses' => 'AccountController@index']);
    Route::get('connect', ['as' => 'connect', 'uses' = > 'AccountController@connect']);
});

Look here 用于解释并在 official documentation（在 Route Groups & Named Routes 下）。

更新

{{ $routeName = \Request::route()->getName() }}

@if(strpos($routeName, 'account.') === 0)
    // do something
@endif

Rohit Khatri 的替代品

function getCurrentRouteGroup() {
    $routeName = Illuminate\Support\Facades\Route::current()->getName();
    return explode('.',$routeName)[0];
}

Answer 2

试试这个

Route::group(['prefix'=>'accounts','as'=>'account.'], function(){

Route::get('connect', [
'as' => 'connect', 'uses' => 'AccountController@connect'
]);

});

Answer 3

// both the format of defining the prefix are working,tested on laravel 5.6

Route::group(['prefix'=>'accounts','as'=>'account.'], function() {
    Route::get('/', 'SomeController@index')->name('test');
    Route::get('/new', function(){
            return redirect()->route('account.test');
    });
});

Route::group(['prefix' => 'admin', 'as' => 'admin.'], function () {
    Route::get('/', [
        'as' => 'custom',
        'uses' => 'SomeController@index'
    ]);  

    Route::get('/custom', function(){
        return route('admin.custom');
    });
}); 

Answer 4

它应该可以工作-

刀片内部-

{{ $yourRouteName = \Request::route()->getName() }}

// Find the first occurrence of account in URL-

@if(strpos($routeName, 'account.') === 0)
  console the message or your code
@endif

Answer 5

您可以使用Route::name()->group(...) 为一组路由的所有名称添加前缀

Route::name('foo.')->prefix('xyz')->group(function() {

    Route::get('path', 'SomeController@method')->name('bar');

});

这里 route('foo.bar') 解析为 url /xyz/path

查看相关Laravel Docs

不要忘记在前缀名称中添加点:-)

Answer 6

在 Laravel 9 中，您现在可以这样做：

Route::controller(AccountController::class)->group(function () {

Route::get('/', 'index')->name('index');
Route::get('/connect', 'connect')->name('connect');

});