Laravel 5 在laravel 5.7中对null调用名为()的成员函数
我有这条路线Laravel 5 在laravel 5.7中对null调用名为()的成员函数,laravel-5,routes,Laravel 5,Routes,我有这条路线 Route::get('/books/science', 'BookController@science')->name('scientific'); 当我想根据文件检查此条件时: 我犯了这个错误。当我添加路线时也是如此 use Illuminate\Support\Facades\Route; 然后扔掉这些代码 dd(Route::currentRouteName()); dd(\Request::route()); 我得到空值$request->route将在全局中
Route::get('/books/science', 'BookController@science')->name('scientific');
当我想根据文件检查此条件时:
我犯了这个错误。当我添加路线时也是如此
use Illuminate\Support\Facades\Route;
然后扔掉这些代码
dd(Route::currentRouteName());
dd(\Request::route());
我得到空值$request->route将在全局中间件中返回空值
将中间件添加到其他组中,如下所示:
protected $middlewareGroups = [
'web' => [
...
YOUR_MIDDLEWARE::class,
]
]
或者将其添加到$RouteMiddle软件中
并将其应用于您的路线
Route::middleware('your_middleware')->group(function () {
Route::get('/books/science', 'BookController@science')->name('scientific');
});
handle方法在哪里?如何调用它?文档中的示例特定于中间件。是。它位于中间件中,中间件在kernel.php的受保护$Middleware中注册
protected $routeMiddleware = [
...
'your_middleware' => YOUR_MIDDLEWARE::class,
];
Route::middleware('your_middleware')->group(function () {
Route::get('/books/science', 'BookController@science')->name('scientific');
});