CODEWITHSUNDEEP
phplaravellaravel-5.1
scott
scott

Reputation: 3202

Show a 404 page if route not found in Laravel 5.1

I am trying to figure out to show 404 page not found if a route is not found. I followed many tutorials, but it doesn't work. I have 404.blade.php in \laravel\resources\views\errors

Also in handler.php

public function render($request, Exception $e)
{
    if ($e instanceof TokenMismatchException) {
        // redirect to form an example of how i handle mine
        return redirect($request->fullUrl())->with(
            'csrf_error',
            "Opps! Seems you couldn't submit form for a longtime. Please try again"
        );
    }

    /*if ($e instanceof CustomException) {
        return response()->view('errors.404', [], 500);
    }*/

    if ($e instanceof \Symfony\Component\HttpKernel\Exception\NotFoundHttpException)
        return response(view('error.404'), 404);

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

If I enter wrong URL in browser, it returns a blank page. I have 

'debug' => env('APP_DEBUG', true),

in app.php.

Can anyone help me how to show a 404 page if route is not found? Thank you.

Upvotes: 23

Views: 67649

Answers (8)

Sarfraz Rasheed
Sarfraz Rasheed

Reputation: 121

If you want to redirect to the 404 page if the route is not found..

Route::fallback(function () {
    return view('404');
});

Upvotes: 0

hassan javaid
hassan javaid

Reputation: 82

In config folder app.php change the following code

'debug' => env('APP_DEBUG', true),

In App->Exception->Handler.php Replace Render Function With Below Code

public function render($request, Exception $exception)
{
    
    if ($exception instanceof ModelNotFoundException) 
    {
    return response()->view('errors.404', [], 404);
    }

    
    if ($exception instanceof \ErrorException) {
    return response()->view('errors.500', [], 500);
    } 
    else {
    return parent::render($request, $exception);
    }
    return parent::render($request, $exception);
}

Upvotes: 0

user4294557
user4294557

Reputation:

@tester.Your problem has already been solved, try the command below in composer:

php artisan view:clear

Then try once more with an unknown URL. Because I have also faced the same error before.

Upvotes: 6

Pim
Pim

Reputation: 5906

I recieved 500 errors instead of 404 errors. I solved the problem like this:

In the app/Exceptions/Handler.php file, there is a render function.

Replace the function with this function:

public function render($request, Exception $e)
{
    if ($this->isHttpException($e)) {
        switch ($e->getStatusCode()) {

            // not authorized
            case '403':
                return \Response::view('errors.403',array(),403);
                break;

            // not found
            case '404':
                return \Response::view('errors.404',array(),404);
                break;

            // internal error
            case '500':
                return \Response::view('errors.500',array(),500);
                break;

            default:
                return $this->renderHttpException($e);
                break;
        }
    } else {
        return parent::render($request, $e);
    }
}

You can then use views that you save in views/errors/404.blade.php, and so on.

Upvotes: 30

Shinseiki86
Shinseiki86

Reputation: 11

I use the following in app/Exceptions/Handler.php (Laravel 5.2):

/**
 * Render an exception into an HTTP response.
 *
 * @param  \Illuminate\Http\Request  $request
 * @param  \Exception  $e
 * @return \Illuminate\Http\Response
 */
public function render($request, Exception $e)
{
    if ($e instanceof \ReflectionException OR $e instanceof \Symfony\Component\HttpKernel\Exception\NotFoundHttpException) //Si la ruta no existe, mostar view 404.
        return response(view('errors.404'), 404);
    return parent::render($request, $e);
}

And it looks like this:img

Upvotes: 1

Mohammed Elhag
Mohammed Elhag

Reputation: 4302

In apache you could be able to put this code in .htaccess file at your main directory and make sure that change AllowOverride Directive to all in httpd confg file

ErrorDocument 404 the\path\to\404.blade.php

Upvotes: 0

patricus
patricus

Reputation: 62228

There is no need for you to check the error type and manually render the 404 view. Laravel already knows to render the view with the HTTP error code that was thrown (404 = resources/views/errors/404.blade.php). Get rid of the extra check and it should work fine.

public function render($request, Exception $e)
{
    if ($e instanceof TokenMismatchException) {
        // redirect to form an example of how i handle mine
        return redirect($request->fullUrl())->with(
            'csrf_error',
            "Opps! Seems you couldn't submit form for a longtime. Please try again"
        );
    }

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

Upvotes: 1

mdamia
mdamia

Reputation: 4557

enter image description here> The abort method will immediately raise an exception which will be rendered by the exception handler. Optionally, you may provide the response text:

abort(403, 'Unauthorized action.');

is your app_debug set to true? if that is the case, Laravel will throw the error with backtrace for debugging purposes, if you change the value to false, Laravel will show the default 404 page in the errors folder. That being said you can choose to use abort at any time you want. at the controller level or at the route level, it is totally up to you.

ie 

Route::get('/page/not/found',function($closure){
  // second parameter is optional. 
  abort(404,'Page not found');
  abort(403); 
});

Upvotes: 9

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