I am new to laravel and want to handle exception if a user hits a urls by using wrong HTTP method. I want to send user response in json but code is not working, sometimes it gives blanks page. Below is my code:
Handler.php
public function render($request, Exception $exception)
{
if ($exception instanceof MethodNotAllowedHttpException) {
return response()->json(['error' => 'Bad Request.'], 404);
}
}
Thanks in advance
try this:
public function render($request, Exception $exception)
{
if (basename(str_replace('\\', '/', get_class($exception))) == 'MethodNotAllowedHttpException') {
return response()->json(['error' => 'Bad Request.'], 404);
}
return parent::render($request, $exception);
}
Related
I want to return a JSON response instead of the default 404 error page when ModelNotFoundException occurs. To do this, I wrote the following code into app\Exceptions\Handler.php :
public function render($request, Exception $exception)
{
if ($exception instanceof ModelNotFoundException) {
return response()->json([
'error' => 'Resource not found'
], 404);
}
return parent::render($request, $exception);
}
However it doesn't work. When the ModelNotFoundException occurs, Laravel just shows a blank page. I find out that even declaring an empty render function in Handler.php makes Laravel display a blank page on ModelNotFoundException.
How can I fix this so it can return JSON/execute the logic inside the overriden render function?
In Laravel 8x, You need to Rendering Exceptions in register() method
use App\Exceptions\CustomException;
/**
* Register the exception handling callbacks for the application.
*
* #return void
*/
public function register()
{
$this->renderable(function (CustomException $e, $request) {
return response()->view('errors.custom', [], 500);
});
}
For ModelNotFoundException you can do it as below.
use Symfony\Component\HttpKernel\Exception\NotFoundHttpException;
public function register()
{
$this->renderable(function (NotFoundHttpException $e, $request) {
return response()->json(...);
});
}
By default, the Laravel exception handler will convert exceptions into an HTTP response for you. However, you are free to register a custom rendering Closure for exceptions of a given type. You may accomplish this via the renderable method of your exception handler. Laravel will deduce what type of exception the Closure renders by examining the type-hint of the Closure:
More info about the error exception
This code doesn't work for me (in Laravel 8.74.0):
$this->renderable(function (ModelNotFoundException$e, $request) {
return response()->json(...);
});
Don't know why, but ModelNotFoundException is directly forwarded to NotFoundHttpException (which is a part of Symfony Component) that used by Laravel and will ultimately triggers a 404 HTTP response. My workaround is checking the getPrevious() method of the exception:
$this->renderable(function (NotFoundHttpException $e, $request) {
if ($request->is('api/*')) {
if ($e->getPrevious() instanceof ModelNotFoundException) {
return response()->json([
'status' => 204,
'message' => 'Data not found'
], 200);
}
return response()->json([
'status' => 404,
'message' => 'Target not found'
], 404);
}
});
And then we will know that this exception come from ModelNotFoundException and return a different response with NotFoundHttpException.
Edit
This is why ModelNotFoundException thrown as NotFoundHttpException
This one is my Handler file:
use Throwable;
public function render($request, Throwable $exception)
{
if( $request->is('api/*')){
if ($exception instanceof ModelNotFoundException) {
$model = strtolower(class_basename($exception->getModel()));
return response()->json([
'error' => 'Model not found'
], 404);
}
if ($exception instanceof NotFoundHttpException) {
return response()->json([
'error' => 'Resource not found'
], 404);
}
}
}
This one is only for all request in API route. If you want to catch all request, so remove the first if.
Please note that by default Laravel emits a JSON representation of an exception ONLY when you send a request with the header parameter Accept: application/json! For all other requests, Laravel sends normal HTML rendered output.
I want to return a json response when an api call is made to a laravel 5.7 app api route when the model is not found. To do this I have modified the render() method of app\Exceptions\Handler.php like this
public function render($request, Exception $exception)
{
if ($exception instanceof ModelNotFoundException && $request->wantsJson()) {
return response()->json(['message' => 'Not Found!'], 404);
}
return parent::render($request, $exception);
}
and my controller show() method is using a Book model like this
public function show(Book $book)
{
return new BookResource($book->load('ratings'));
}
Test on postman, a get call to localhost:8000/api/books/1 (id 1 has been deleted) keeps returning the default laravel 404 not found page instead of json.
Have I missed a step or something? I also noticed that adding a conditional statement inside the controller show() method like this
public function show(Book $book)
{
if ($book) {
return new BookResource($book->load('ratings'));
} else {
return response()->json(['message' => 'Not found'], 404);
}
}
returns the same html result instead of json.
What will be the proper way to handle this scenario?
Your code is correct. The problem is that you are probably testing it on a Local environment so in your .env you have set:
APP_DEBUG=true, switch it to APP_DEBUG=false and you will see your custom message.
PS: $request->wantsJson() is not necessary if your clients send the correct header info, eg: 'accept:application/json'
You can remove $request->wantsJson
or you can set the header in your request "Accept" => "application/json"
May this can help you:
public function render($request, Exception $exception)
{
if ($exception instanceof ModelNotFoundException && ($request->wantsJson() || $request->ajax())) {
return response()->json(['message' => 'Not Found!'], 404);
}
return parent::render($request, $exception);
}
I am working on a RESTful application using Laravel 5 and I am trying to catch exceptions and generate an appropriate response. I am also using the tymondesigns/jwt-auth package so that all the API responses are in JSend JSON format.
Right now I am trying to catch the TokenExpiredException which arises when the given token is expired of course. So I tried this in the Handler.php:
if($e instanceof TokenExpiredException)
{
return jsend()->error()
->message("Token Expired")
->code(403)
->data([null])
->get();
}
But I am still not able to catch this exception and give back a JSON response. Although I am able to do this for other exceptions like:
if ($e instanceof ModelNotFoundException) {
$e = new NotFoundHttpException($e->getMessage(), $e);
return jsend()->error()
->message("404 Model Not Found")
->data([null])
->get();
}
And:
if ($this->isHttpException($e))
{
if($e instanceof NotFoundHttpException)
{
return jsend()->error()
->message("404 Route Not Found")
->data([null])
->get();
}
return $this->renderHttpException($e);
}
How to handle other exceptions in Laravel?
It seems I forgot to use the namespace:
if($e instanceof \Tymon\JWTAuth\Exceptions\TokenExpiredException)
{
return jsend()->error()
->message("Token Expired")
->code(403)
->data([null])
->get();
}
Small mistakes! facepalm
If someone comes wondering here with same problem for new Laravel (5.4) and jwt-auth (1.0.*#dev)... now there is another cause/solution to this.
Provider catches instance of \Tymon\JWTAuth\Exceptions\TokenExpiredException and rethrows instance of Symfony\Component\HttpKernel\Exception\UnauthorizedHttpException. Original exception is still available with method getPrevious(), so error handling would now look something like this:
public function render($request, Exception $exception)
{
if ($exception->getPrevious() instanceof \Tymon\JWTAuth\Exceptions\TokenExpiredException) {
return response()->json(['error' => $exception->getPrevious()->getMessage()], $exception->getStatusCode());
} else if ($exception->getPrevious() instanceof \Tymon\JWTAuth\Exceptions\TokenInvalidException) {
return response()->json(['error' => $exception->getPrevious()->getMessage()], $exception->getStatusCode());
}
return parent::render($request, $exception);
}
I make ajax requests to Laravel backend.
In backend I check request data and throw some exceptions.
Laravel, by default, generate html pages with exception messages.
I want to respond just raw exception message not any html.
->getMessage() doesn't work. Laravel, as always, generate html.
What shoud I do?
In Laravel 5 you can catch exceptions by editing the render method in app/Exceptions/Handler.php.
If you want to catch exceptions for all AJAX requests you can do this:
public function render($request, Exception $e)
{
if ($request->ajax()) {
return response()->json(['message' => $e->getMessage()]);
}
return parent::render($request, $e);
}
This will be applied to ANY exception in AJAX requests. If your app is sending out an exception of App\Exceptions\MyOwnException, you check for that instance instead.
public function render($request, Exception $e)
{
if ($e instanceof \App\Exceptions\MyOwnException) {
return response()->json(['message' => $e->getMessage()]);
}
return parent::render($request, $e);
}
Im just move to laravel 5 and im receiving errors from laravel in HTML page. Something like this:
Sorry, the page you are looking for could not be found.
1/1
NotFoundHttpException in Application.php line 756:
Persona no existe
in Application.php line 756
at Application->abort('404', 'Person doesnt exists', array()) in helpers.php line
When i work with laravel 4 all works fine, the errors are in json format, that way i could parse the error message and show a message to the user. An example of json error:
{"error":{
"type":"Symfony\\Component\\HttpKernel\\Exception\\NotFoundHttpException",
"message":"Person doesnt exist",
"file":"C:\\xampp\\htdocs\\backend1\\bootstrap\\compiled.php",
"line":768}}
How can i achieve that in laravel 5.
Sorry for my bad english, thanks a lot.
I came here earlier searching for how to throw json exceptions anywhere in Laravel and the answer set me on the correct path. For anyone that finds this searching for a similar solution, here's how I implemented app-wide:
Add this code to the render method of app/Exceptions/Handler.php
if ($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 :-)
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, $error->getStatusCode());
}
Laravel 5 offers an Exception Handler in app/Exceptions/Handler.php. The render method can be used to render specific exceptions differently, i.e.
public function render($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.
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);
}
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, $exception);
}
Instead of
if ($request->ajax() || $request->wantsJson()) {...}
use
if ($request->expectsJson()) {...}
vendor\laravel\framework\src\Illuminate\Http\Concerns\InteractsWithContentTypes.php:42
public function expectsJson()
{
return ($this->ajax() && ! $this->pjax()) || $this->wantsJson();
}
I updated my app/Exceptions/Handler.php to catch HTTP Exceptions that were not validation errors:
public function render($request, Exception $exception)
{
// converts errors to JSON when required and when not a validation error
if ($request->expectsJson() && method_exists($exception, 'getStatusCode')) {
$message = $exception->getMessage();
if (is_object($message)) {
$message = $message->toArray();
}
return response()->json([
'errors' => array_wrap($message)
], $exception->getStatusCode());
}
return parent::render($request, $exception);
}
By checking for the method getStatusCode(), you can tell if the exception can successfully be coerced to JSON.
If you want to get Exception errors in json format then
open the Handler class at App\Exceptions\Handler and customize it.
Here's an example for Unauthorized requests and Not found responses
public function render($request, Exception $exception)
{
if ($exception instanceof AuthorizationException) {
return response()->json(['error' => $exception->getMessage()], 403);
}
if ($exception instanceof ModelNotFoundException) {
return response()->json(['error' => $exception->getMessage()], 404);
}
return parent::render($request, $exception);
}