标签: laravel-middleware

我在Laravel应用程序中试验中间件.我目前已将其设置为在经过身份验证的用户的每个路由上运行,但是,我希望它忽略以setupURI 开头的任何请求.

这是我的CheckOnboarding中间件方法的样子:

public function handle($request, Closure $next)
{
    /** 
    * Check to see if the user has completed the onboarding, if not redirect.
    * Also checks that the requested URI isn't the setup route to ensure there isn't a redirect loop.
    */
    if ($request->user()->onboarding_complete == false && $request->path() != 'setup') {
        return redirect('setup');
    } else {
        return $next($request);
    }
}

这在我的路线中使用如下:

Route::group(['middleware' => ['auth','checkOnboarding']], function () {
    Route::get('/home', 'HomeController@index');
    Route::get('/account', 'AccountController@index');

    Route::group(['prefix' => 'setup'], function …

我试图在我的条件下给予条件Middleware.

这是我的剧本

if (auth()->check() && auth()->user()->type == 'TP001') {
    $menu->add("User Control",array('nickname' => "user",'class'=>'treeview'))
    ->append(' <b class="caret"></b>')
    ->prepend('<span class="glyphicon glyphicon-user"></span> ');

    $menu->user->add('Daftar User','user/list');
    $menu->user->add('Tipe User','user/type');
} else {
    /* Some code here...*/
}

上面的脚本我不能看到带有条件的菜单,即使我已经登录'TP001'(总是在其他地方),然后我尝试修复我的代码

auth()->user()->isDeveloper()

我的模特

public function isDeveloper()
{
    return ($this->type == 'TP001');
}

但仍然没有工作,有没有办法给出上述条件但是以正确的方式？提前谢谢,抱歉我的英语不好.

我的核心

  protected $middleware = [
        \Illuminate\Foundation\Http\Middleware\CheckForMaintenanceMode::class,
        \Illuminate\Foundation\Http\Middleware\ValidatePostSize::class,
        \App\Http\Middleware\TrimStrings::class,
        \Illuminate\Foundation\Http\Middleware\ConvertEmptyStringsToNull::class,
        \App\Http\Middleware\Frontend::class,
  ];

某些服务向我的站点发出 HTTP 请求并传递一些输入。这个输入的结构对我来说有点错误，所以我正在尝试修改它。

我制作了一个中间件并将这个中间件附加到我的路线上。句柄方法如下所示：

public function handle($request, Closure $next)
{
    $input = $request->all();

    // Input modification

    $request->replace($input);
    \Log::info($request->all()); // Shows modified request

    return $next($request);
}

但是在我的控制器中，我得到了旧的输入。另外我有点困惑，因为我也使用 FormRequest，而且我意识到这两个请求是不同的实体。那么如何修改中间件中的输入呢？

----------------------------------------------------------------------------------+
| Domain | Method   | URI                        | Name             | Action                                                                 | Middleware
                                                                                  |
+--------+----------+----------------------------+------------------+------------------------------------------------------------------------+--------------------------------------------------
----------------------------------------------------------------------------------+
|        | GET|HEAD | /                          |                  | Closure                                                                | web
                                                                                  |
|        | POST     | _ignition/execute-solution |                  | Facade\Ignition\Http\Controllers\ExecuteSolutionController             | Facade\Ignition\Http\Middleware\IgnitionEnabled,F
acade\Ignition\Http\Middleware\IgnitionConfigValueEnabled:enableRunnableSolutions |
|        | GET|HEAD | _ignition/health-check     |                  | Facade\Ignition\Http\Controllers\HealthCheckController                 | Facade\Ignition\Http\Middleware\IgnitionEnabled
                                                                                  |
|        | GET|HEAD | _ignition/scripts/{script} |                  | Facade\Ignition\Http\Controllers\ScriptController                      | Facade\Ignition\Http\Middleware\IgnitionEnabled
                                                                                  |
|        | POST     | _ignition/share-report     |                  | Facade\Ignition\Http\Controllers\ShareReportController                 | Facade\Ignition\Http\Middleware\IgnitionEnabled,F
acade\Ignition\Http\Middleware\IgnitionConfigValueEnabled:enableShareButton       |
|        | GET|HEAD | _ignition/styles/{style}   |                  | …

目标

我正在尝试为登录用户创建管理员路由限制.我已经尝试检查我的用户是否log-in,以及用户类型是否Admin,如果是,我想允许他们访问管理路由,否则,回复404.

routes.php文件

<!-- Route group -->
$router->group(['middleware' => 'auth'], function() {


    <!-- No Restriction -->
    Route::get('dashboard','WelcomeController@index');

    <!-- Admin Only -->
    if(Auth::check()){
        if ( Auth::user()->type == "Admin" ){

            //Report
            Route::get('report','ReportController@index');
            Route::get('report/create', array('as'=>'report.create', 'uses'=>'ReportController@create'));
            Route::post('report/store','ReportController@store');
            Route::get('report/{id}', array('before' =>'profile', 'uses'=>'ReportController@show'));
            Route::get('report/{id}/edit', 'ReportController@edit');
            Route::put('report/{id}/update', array('as'=>'report.update', 'uses'=>'ReportController@update'));
            Route::delete('report/{id}/destroy',array('as'=>'report.destroy', 'uses'=>'ReportController@destroy'));

        }
    }

});

结果

它没有像我预期的那样工作.它会引发404错误 - 即使对于Admin用户也是如此.

我正在使用Auth脚手架Laravel 5.3,我已经改变了路线auth.因此,而不是/login和/register我使用/signin和/signup.

在Laravel 5.2我们在有这个默认auth的中间件,

public function handle($request, Closure $next, $guard = null)
{
    if (Auth::guard($guard)->guest()) {
        if ($request->ajax() || $request->wantsJson()) {
            return response('Unauthorized.', 401);
        }

        return redirect()->guest('login');
    }

    return $next($request);
}

login如果用户没有登录,这将重定向到路由.Laravel 5.3我们有这个,

public function handle($request, Closure $next, $guard = null)
{
    if (Auth::guard($guard)->check()) {
        return redirect('/');
    }

    return $next($request);
}

这会将已登录的用户重定向到默认路由/.所以他们在5.3中改变了它.而不是定义客人去哪里,我们定义了登录用户去.

我的问题是,我将如何原 …

Laravel 5.7

PHP 7.2.10

目前我可以使用 web 和 api 守卫中的任何一个，有什么方法可以同时使用两者，以便 web 应用程序和 api 可以一起工作。

就像是

return [

    /*
    |--------------------------------------------------------------------------
    | Authentication Defaults
    |--------------------------------------------------------------------------
    |
    | This option controls the default authentication "guard" and password
    | reset options for your application. You may change these defaults
    | as required, but they're a perfect start for most applications.
    |
    */

    'defaults' => [
        'guard' => 'api|web',
        'passwords' => 'users',
    ],

如果不使用架构，这是一个需要更改架构的解决方案/解决方法，这是我不喜欢的。此外，我不需要访问令牌进行注册，这个答案在做什么。

api.php

Route::group([
    'middleware' => 'api|web',
    'prefix' => 'auth'
], function …

你好开发者,

这不是直接重复:调用成员函数setCookie()on null 当我从Middleware调用Controller操作然后在控制器操作中我返回View时,

这是我的路线 web.php

Route::get('/end', [EndPageController::class, 'index'])
->name('survey.end')
->middleware('App\Http\Middleware\LegacyEndCheck');

中间件LegacyEndCheck.php处理功能

public function handle($request, Closure $next)
{
    $sjid = $request->input('sjid', false);

    if( empty($sjid) || strlen($sjid) !== 36 ){
        return app()->make(EndLegacyController::class)->index($request); //I have to call EndLegacyController@index here as I can't change URL using redirect route
    }

    return $next($request);
}

控制器EndLegacyController index动作

public function index(Request $request)
{
    $sjid = $request->input('sjid',false);
    $status = $request->input('status',false);
    if( empty($sjid) || empty($status) ){
        $this->status = 5;
        return view('survey.legacy.end')
            ->with('status', $this->status);
    } …

在我的应用程序中,我实现了OAuth2-Server(oauth2-server-laravel)与自定义身份验证包(Sentinel by Cartalyst)的组合.

在我的routes.php中:

Route::group(['before' => 'oauth'], function()
{
    // ... some routes here
}

因此,请求必须提供授权标头,否则应用程序将退出OAuthException.

现在我想对我的控制器进行单元测试.因此,我必须为每个测试使用OAuth会话和访问令牌为我的数据库建立种子.然后覆盖call()-method TestCase并使用Bearer Token设置HTTP-Authorization Header.

有没有办法禁用或绕过中间件(在我的情况下仅用于单元测试)？

在Laravel 4中,它们被称为路由过滤器,无论如何它们在测试环境中被禁用.您也可以手动启用/禁用它们Route::enableFilters().

我用laravel后端制作了反应项目......我有一个CORS问题，我像下面的链接一样做所有事情，用水果蛋糕。

        'paths' => ['api/*'],

    /*
    * Matches the request method. `[*]` allows all methods.
    */
    'allowed_methods' => ['*'],

    /*
     * Matches the request origin. `[*]` allows all origins.
     */
    'allowed_origins' => ['*'],

    /*
     * Matches the request origin with, similar to `Request::is()`
     */
    'allowed_origins_patterns' => [],

    /*
     * Sets the Access-Control-Allow-Headers response header. `[*]` allows all headers.
     */
    'allowed_headers' => ['*'],

    /*
     * Sets the Access-Control-Expose-Headers …