Arrays 将带有资源路由的参数传递给控制器laravel
我有这样的资源路线Arrays 将带有资源路由的参数传递给控制器laravel,arrays,laravel,controller,routes,resources,Arrays,Laravel,Controller,Routes,Resources,我有这样的资源路线 Route::resource('users','UsersController'); Route::resource('users','UsersController')-> with array to use it in the functions; // with array passed 我想把数组和变量传递给索引页面或创建页面 所以我需要这样 Route::resource('users','UsersController'); Route::reso
Route::resource('users','UsersController');
Route::resource('users','UsersController')-> with array to use it in the functions; // with array passed
我想把数组和变量传递给索引页面或创建页面
所以我需要这样
Route::resource('users','UsersController');
Route::resource('users','UsersController')-> with array to use it in the functions; // with array passed
像这样在我的控制器中使用它
public function index()
{
return $my_array_passed_from_route;
if(Gate::allows('users.view'))
{
$users = User::withTrashed()->paginate(100);
return view('users.index',compact('users'));
}
else
return Helper::not_auth();
}
非常感谢你一定要试试这个
Route::get('users/{id}', [
'as' => 'users.show',
'uses' => 'UsersController@show'
]);
Route::resource('user', 'UsersController', ['except' => 'show']);
你必须试试这个
Route::get('users/{id}', [
'as' => 'users.show',
'uses' => 'UsersController@show'
]);
Route::resource('user', 'UsersController', ['except' => 'show']);
也许你能帮助我。看看这个:也许你能帮我。看看这个: