How to send Laravel error responses as JSON

Question

Im just move to laravel 5 and im receiving errors from laravel in HTML page. Something like this:

Sorry, the page you are looking for could not be found.

1/1
NotFoundHttpException in Application.php line 756:
Persona no existe
in Application.php line 756
at Application->abort('404', 'Person doesnt exists', array()) in helpers.php line 

When i work with laravel 4 all works fine, the errors are in json format, that way i could parse the error message and show a message to the user. An example of json error:

{"error":{
"type":"Symfony\\Component\\HttpKernel\\Exception\\NotFoundHttpException",
"message":"Person doesnt exist",
"file":"C:\\xampp\\htdocs\\backend1\\bootstrap\\compiled.php",
"line":768}}

How can i achieve that in laravel 5.

Sorry for my bad english, thanks a lot.

Answer 1

I came here earlier searching for how to throw json exceptions anywhere in Laravel and the answer set me on the correct path. For anyone that finds this searching for a similar solution, here's how I implemented app-wide:

Add this code to the render method of app/Exceptions/Handler.php

if ($request->ajax() || $request->wantsJson()) {
    return new JsonResponse($e->getMessage(), 422);
}

Add this to the method to handle objects:

if ($request->ajax() || $request->wantsJson()) {

    $message = $e->getMessage();
    if (is_object($message)) { $message = $message->toArray(); }

    return new JsonResponse($message, 422);
}

And then use this generic bit of code anywhere you want:

throw new \Exception("Custom error message", 422);

And it will convert all errors thrown after an ajax request to Json exceptions ready to be used any which way you want :-)

Answer 2

Laravel 5.1

To keep my HTTP status code on unexpected exceptions, like 404, 500 403...

This is what I use (app/Exceptions/Handler.php):

 public function render($request, Exception $e)
{
    $error = $this->convertExceptionToResponse($e);
    $response = [];
    if($error->getStatusCode() == 500) {
        $response['error'] = $e->getMessage();
        if(Config::get('app.debug')) {
            $response['trace'] = $e->getTraceAsString();
            $response['code'] = $e->getCode();
        }
    }
    return response()->json($response, $error->getStatusCode());
}

Answer 3

Laravel 5 offers an Exception Handler in app/Exceptions/Handler.php. The render method can be used to render specific exceptions differently, i.e.

public function render($request, Exception $e)
{
    if ($e instanceof API\APIError)
        return \Response::json(['code' => '...', 'msg' => '...']);
    return parent::render($request, $e);
}

Personally, I use App\Exceptions\API\APIError as a general exception to throw when I want to return an API error. Instead, you could just check if the request is AJAX (if ($request->ajax())) but I think explicitly setting an API exception seems cleaner because you can extend the APIError class and add whatever functions you need.

Answer 4

Edit: Laravel 5.6 handles it very well without any change need, just be sure you are sending Accept header as application/json.

---

If you want to keep status code (it will be useful for front-end side to understand error type) I suggest to use this in your app/Exceptions/Handler.php:

public function render($request, Exception $exception)
{
    if ($request->ajax() || $request->wantsJson()) {

        // this part is from render function in Illuminate\Foundation\Exceptions\Handler.php
        // works well for json
        $exception = $this->prepareException($exception);

        if ($exception instanceof \Illuminate\Http\Exception\HttpResponseException) {
            return $exception->getResponse();
        } elseif ($exception instanceof \Illuminate\Auth\AuthenticationException) {
            return $this->unauthenticated($request, $exception);
        } elseif ($exception instanceof \Illuminate\Validation\ValidationException) {
            return $this->convertValidationExceptionToResponse($exception, $request);
        }

        // we prepare custom response for other situation such as modelnotfound
        $response = [];
        $response['error'] = $exception->getMessage();

        if(config('app.debug')) {
            $response['trace'] = $exception->getTrace();
            $response['code'] = $exception->getCode();
        }

        // we look for assigned status code if there isn't we assign 500
        $statusCode = method_exists($exception, 'getStatusCode') 
                        ? $exception->getStatusCode()
                        : 500;

        return response()->json($response, $statusCode);
    }

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

Answer 5

On Laravel 5.5, you can use prepareJsonResponse method in app/Exceptions/Handler.php that will force response as JSON.

/**
 * Render an exception into an HTTP response.
 *
 * @param  \Illuminate\Http\Request  $request
 * @param  \Exception  $exception
 * @return \Illuminate\Http\Response
 */
public function render($request, Exception $exception)
{
    return $this->prepareJsonResponse($request, $exception);
}