Question

在Laravel 5中，是否有办法在应用程序内部/编程方式调用路由？我已经为Laravel 4找到了很多教程，但我找不到版本5的信息。

Answer 1

使用laravel 5.5，这种方法对我有用：

$req = Request::create('/my/url', 'POST', $params);
$res = app()->handle($req);
$responseBody = $res->getContent();
// or if you want the response to be json format  
// $responseBody = json_decode($res->getContent(), true);

来源： https://laracasts.com/discuss/channels/laravel/route-dispatch

*注意：如果您尝试访问的路线可能会有问题有身份验证中间件，你没有提供正确的凭据。为避免这种情况，请务必设置所需的正确标头，以便正常处理请求（例如Authorisation bearer ...）。

Answer 2

您可以尝试这样的事情：

// GET Request
$request = Request::create('/some/url/1', 'GET');
$response = Route::dispatch($request);

// POST Request
$request = Request::create('/some/url/1', 'POST', Request::all());
$response = Route::dispatch($request);

Answer 3

您实际上可以呼叫与该路由关联的控制器，而不是“呼叫”＃39;内部路线。

例如：

<强> routes.php文件

Route::get('/getUser', 'UserController@getUser');

<强> UserController.php

class UserController extends Controller {

   public function getUser($id){

      return \App\User::find($id);

   };
}

您可以实际拨打/getUser，而不是拨打UserController@getUser路由。

$ctrl = new \App\Http\Controllers\UserController();
$ctrl->getUser(1);

如果你的意思是，这与内部路线的calling相同。希望有所帮助

Answer 4

// this code based on laravel 5.8
// I tried to solve this using guzzle first . but i found guzzle cant help me while I 
//am using same port. so below is the answer
// you may pass your params and other authentication related data while calling the 
//end point
public function profile(){

//    '/api/user/1' is my api end please put your one
// 
$req = Request::create('/api/user/1', 'GET',[ // you may pass this without this array
        'HTTP_Accept' => 'application/json', 
        'Content-type' => 'application/json'
        ]);
$res = app()->handle($req);
$responseBody = json_decode($res->getContent()); // convert to json object using 
json_decode and used getcontent() for getting content from response 
return response()->json(['msg' =>$responseBody ], 200); // return json data with 
//status code 200
   }

Answer 5

这些答案都不适合我：他们要么不接受查询参数，要么不能使用现有的app（）实例（config＆amp; .env vars需要）。

我想在内部调用路由，因为我正在编写控制台命令来与我的应用程序API进行交互。

这是我做的对我有用的事情：

<?php // We're using Laravel 5.3 here.

namespace App\Console;

use App\MyModel;
use App\MyOtherModel;
use App\Http\Controllers\MyController;
use Illuminate\Console\Command;

class MyCommand extends Command
{
    protected $signature = 'mycommand
                            {variable1} : First variable
                            {variable2} : Another variable';

    public function handle()
    {
        // Set any required headers. I'm spoofing an AJAX request:
        request()->headers->set('X-Requested-With', 'XMLHttpRequest');

        // Set your query data for the route:
        request()->merge([
            'variable1' => $this->argument('variable1'),
            'variable2'  => $this->argument('variable2'),
        ]);

        // Instantiate your controller and its dependencies:
        $response = (new MyController)->put(new MyModel, new MyOtherModel);

        // Do whatever you want with the response:
        var_dump($response->getStatusCode()); // 200, 404, etc.
        var_dump($response->getContent()); // Entire response body

        // See what other fun stuff you can do!:
        var_dump(get_class_methods($response));
    }
}

您的控制器/路由将完全像您使用curl调用它一样工作。玩得开心！

Laravel 5：内部呼叫路由

5 个答案: