TP5.1路由中间件分组的问题

Question

这样访问没有问题 Route::get('/register', 'index/register/register')->middleware('Login'); Route::get('/login', 'index/login/login')->middleware('Login'); 手册上也说了支持路由组的中间件形式 但是改成下面路由组的形式再访问时就报错了 这是怎么回事？ Route::group('/test', function(){ Route::get('/login', 'index/login/login'); Route::get('/register', 'index/register/register'); })->middleware('Login'); 中间件 <?php namespace app\http\middleware; use think\facade\Cookie; use think\response; use think\facade\Request; class Login { public function handle($request, \Closure $next) { if (Cookie::has('uid')) { return redirect('/'); }else{ return redirect('/login'); } return $next($request); } }

Answer 1

Route::group('user', function(){
    Route::get('login', 'index/login/login');
    Route::get('register', 'index/register/register');
})->middleware('Login');

访问/user/login即可

Answer 2

Route::group('test', function(){
    Route::get('test/login', 'index/login/login');
    Route::get('test/register', 'index/register/register');
})->middleware('Login');

试着这样写一下

Answer 3

还有人吗..

Answer 4

谢邀，但是平时不用 ThinPHP, 你可以贴一下运行树，比较好排查

Answer 5

Route::group(['test'], function(){

Route::get('login', 'index/login/login');
Route::get('register', 'index/register/register');

})->middleware('Login');
这样写，我用的5.1.19,欢迎骚扰。。。