Laravel在没有日志记录时引用应用程序的url - Laravel refer to application url when no logged

Laravel refer to application url when no logged

我在下面有路由文件

Route::group(['middleware' => ['auth:web', 'web'],  'prefix' => 'api/1.0'], function () {
    Route::get('dashboard',  'DashboardController@index');
    Route::get('site/{id}', 'SiteController@index');
});

当我没有登录，并试图参考url http://app.url/api/1.0/，是返回登录面板。我如何设置而不是登录面板，返回json数据(例如[message: forbidden])。我尝试在中间件设置，但larver崩溃了。谢谢你

您可以使用auth:api中间件:

Route::group(['middleware' => ['auth:api'],  'prefix' => 'api/1.0'], function () {
    Route::get('dashboard',  'DashboardController@index');
    Route::get('site/{id}', 'SiteController@index');
});

所有异常由App'Exceptions'Handler类处理。您可以编辑该类的render()函数，使其返回JSON响应:

public function render($request, Exception $exception)
{
    if ($request->is('api/*');) {
        return $this->renderRestException($request, $exception);
    }
    return parent::render($request, $exception);
}

在同一个文件中添加两个函数:

/**
 * Render an exception into a response.
 *
 * @param  'Illuminate'Http'Request  $request
 * @param  'Exception  $e
 * @return 'Illuminate'Http'JsonResponse
 */
public function renderRestException(Request $request, Exception $e)
{
    switch($e)
    {
        case ($e instanceof HttpResponseException):
            return response()->json($e->getResponse()->getContent(), $e->getResponse()->getStatusCode());
        case ($e instanceof ModelNotFoundException):
            return response()->json($e->getMessage(), 404);
        case ($e instanceof AuthenticationException):
            return response()->json('Unauthorized', 401);
        case ($e instanceof AuthorizationException):
            return response()->json($e->getMessage(), 403);
        default:
            return $this->convertExceptionToJsonResponse($e);
    }
}
/**
 * Create a Symfony response for the given exception.
 *
 * @param  'Exception  $e
 * @return 'Illuminate'Http'JsonResponse
 */
protected function convertExceptionToJsonResponse(Exception $e)
{
    $e = FlattenException::create($e);
    return response()->json(array_get(SymfonyResponse::$statusTexts, $e->getStatusCode()), $e->getStatusCode());
}

记住要在类的顶部导入所有这些类:

use Exception;
use Illuminate'Http'Request;
use Illuminate'Auth'AuthenticationException;
use Illuminate'Auth'Access'AuthorizationException;
use Illuminate'Http'Exception'HttpResponseException;
use Symfony'Component'Debug'Exception'FlattenException;
use Illuminate'Database'Eloquent'ModelNotFoundException;
use Symfony'Component'HttpFoundation'Response as SymfonyResponse;