如何让 Laravel 返回 JSON REST API 的自定义错误

Question

我正在开发某种 RESTful API。当发生一些错误时，我会抛出一个App::abort($code, $message) 错误。

问题是：我希望他抛出一个带有键“代码”和“消息”的 json 格式的数组，每个都包含上述数据。

Array
(
    [code] => 401
    [message] => "Invalid User"
)

有谁知道这是否可行，如果可行，我该怎么做？

Answer 1

您可以将数组传递给返回的 JSON 响应：

$returnData = array(
    'status' => 'error',
    'message' => 'An error occurred!'
);
return Response::json($returnData, 500);

Answer 2

转到您的app/start/global.php。

这会将 401 和 404 的所有错误转换为自定义 json 错误，而不是 Whoops 堆栈跟踪。添加这个：

App::error(function(Exception $exception, $code)
{
    Log::error($exception);

    $message = $exception->getMessage();

    // switch statements provided in case you need to add
    // additional logic for specific error code.
    switch ($code) {
        case 401:
            return Response::json(array(
                    'code'      =>  401,
                    'message'   =>  $message
                ), 401);
        case 404:
            $message            = (!$message ? $message = 'the requested resource was not found' : $message);
            return Response::json(array(
                    'code'      =>  404,
                    'message'   =>  $message
                ), 404);        
    }

});

这是处理此错误的众多选项之一。

---

制作 API 最好创建自己的帮助程序，例如扩展 Response 类的 Responser::error(400, 'damn')。

有点像：

public static function error($code = 400, $message = null)
{
    // check if $message is object and transforms it into an array
    if (is_object($message)) { $message = $message->toArray(); }

    switch ($code) {
        default:
            $code_message = 'error_occured';
            break;
    }

    $data = array(
            'code'      => $code,
            'message'   => $code_message,
            'data'      => $message
        );

    // return an error
    return Response::json($data, $code);
}

Answer 3

这是我使用的（Laravel 5.2）：

根据：https://laravel.com/docs/5.2/errors，我们可以为app\Exceptions\Handler.php中的错误指定自定义渲染函数。我所做的只是将我的渲染函数更改为：

    /**
     * Render an exception into an HTTP response.
     * Updated to return json for a request that wantsJson 
     * i.e: specifies 
     *      Accept: application/json
     * in its header
     *
     * @param  \Illuminate\Http\Request  $request
     * @param  \Exception  $e
     * @return \Illuminate\Http\Response
     */
    public function render($request, Exception $e)
    {
        if ($request->ajax() || $request->wantsJson()) {
            return response()->json(
                          $this->getJsonMessage($e), 
                          $this->getExceptionHTTPStatusCode($e)
                        );
        }
        return parent::render($request, $e);
    }

    protected function getJsonMessage($e){
        // You may add in the code, but it's duplication
        return [
                  'status' => 'false',
                  'message' => $e->getMessage()
               ];
    }

    protected function getExceptionHTTPStatusCode($e){
        // Not all Exceptions have a http status code
        // We will give Error 500 if none found
        return method_exists($e, 'getStatusCode') ? 
                         $e->getStatusCode() : 500;
    }

在此之后，您需要做的就是确保所有 API 请求都指定 Accept: application/json 标头。希望这会有所帮助:)

Answer 4

根据 Ibrahim 的回答，并非每个 ajax 请求都需要 JSON，响应“状态代码”和“状态”是不必要的，因为它们的含义相同。更重要的是，根本不需要在响应“状态”中提及，因为响应代码“说”了这一点。这样的东西应该可以完美运行：

/**
 * Render an exception into an HTTP response.
 *
 * @param  \Illuminate\Http\Request  $request
 * @param  \Exception  $e
 * @return \Illuminate\Http\Response
 */
public function render($request, Exception $e)
{
    if ($request->wantsJson())
        return response()->json(
            ['message' => $e->getMessage()],
            method_exists($e, 'getStatusCode') ? $e->getStatusCode() : 500);

    return parent::render($request, $e);
}

Answer 5

这是我在 5.6 中使用的，以便返回与内置 validate 方法相同类型的响应：

response()->json(['errors' => ['email' => ['The email is invalid.']]], 422);

Answer 6

在 Laravel5.6 中，我通常为 app\Exceptions\Handler.php 中的错误指定自定义渲染函数。我所做的只是将我的渲染函数更改为：

/**
 * Render an exception into an HTTP response.
 *
 * @param \Illuminate\Http\Request $request
 * @param \Exception               $e
 *
 * @return Response
 */
public function render($request, Exception $e)
{
    if ($request->wantsJson() && !($e instanceof ValidationException)) {
        $response = [
            'message' => (string)$e->getMessage(),
            'status_code' => 400,
        ];

        if ($e instanceof HttpException) {
            $response['message'] = Response::$statusTexts[$e->getStatusCode()];
            $response['status_code'] = $e->getStatusCode();
        } else if ($e instanceof ModelNotFoundException) {
            $response['message'] = Response::$statusTexts[Response::HTTP_NOT_FOUND];
            $response['status_code'] = Response::HTTP_NOT_FOUND;
        }

        if ($this->isDebugMode()) {
            $response['debug'] = [
                'exception' => get_class($e),
                'trace' => $e->getTrace()
            ];
        }

        return response()->json([
            'status'      => 'failed',
            'status_code' => $response['status_code'],
            'massage'     => $response['message'],
        ], $response['status_code']);
    }

    return parent::render($request, $e);
}

Answer 7

Laravel 6：

您必须在客户端的 API 请求中设置 Accept:application/json 标头，Laravel 会自动返回 JSON 格式错误。

 {     
    "message": "Unauthenticated."
 }

Answer 8

$response['message'] ="The given data was invalid";
$error['country_id'] = ["The country field is required"];
$error['state_id'] = ["The state field is required"];
$error['name'] = ["The name field is required"];
$response['error'] = $error;
return response()->json($response,422);

Answer 9

对于 Laravel 8

转到您的 \app\Exceptions\Handler.php 并像这样覆盖 invalidJson 方法：

// Add this line at the top of the class
use Illuminate\Validation\ValidationException;


/**
 * Convert a validation exception into a JSON response.
 *
 * @param  \Illuminate\Http\Request  $request
 * @param  \Illuminate\Validation\ValidationException  $exception
 * @return \Illuminate\Http\JsonResponse
 */
protected function invalidJson($request, ValidationException $exception)
{
    // You can return json response with your custom form
    return response()->json([
        'success' => false,
        'data' => [
            'code' => $exception->status,
            'message' => $exception->getMessage(),
            'errors' => $exception->errors()
        ]
    ], $exception->status);
}

响应样本：

{
  "success": false,
  "data": {
    "code": 422,
    "message": "The given data was invalid.",
    "errors": {
      "password": [
        "The password field is required."
      ]
    }
  }
}

原来的方法是：

/**
 * Convert a validation exception into a JSON response.
 *
 * @param  \Illuminate\Http\Request  $request
 * @param  \Illuminate\Validation\ValidationException  $exception
 * @return \Illuminate\Http\JsonResponse
 */
protected function invalidJson($request, ValidationException $exception)
{
    return response()->json([
        'message' => $exception->getMessage(),
        'errors' => $exception->errors(),
    ], $exception->status);
}

响应样本：

{
  "message": "The given data was invalid.",
  "errors": {
    "password": [
      "The password field is required."
    ]
  }
}

请注意，unauthenticated 响应是在单独的方法中，因此您也可以覆盖它

/**
 * Convert an authentication exception into a response.
 *
 * @param  \Illuminate\Http\Request  $request
 * @param  \Illuminate\Auth\AuthenticationException  $exception
 * @return \Symfony\Component\HttpFoundation\Response
 */
protected function unauthenticated($request, AuthenticationException $exception)
{
    return $request->expectsJson()
        // Here you can change the form of the json response
        ? response()->json(['message' => $exception->getMessage()], 401) // <-
        : redirect()->guest($exception->redirectTo() ?? route('login'));
}