Laravel 5 : How to use findOrFail() method? - php

Question

I just follow some tutorial and so far what I do is :
my App/Exceptions/Handler.php
<?php
...
use Illuminate\Database\Eloquent\ModelNotFoundException;
...
public function render($request, Exception $e)
{
if ($e instanceof ModelNotFoundException){
abort(404);
}
return parent::render($request, $e);
}
and my UsersController looks like this :
...
public function edit($id)
{
$data = User::findOrFail($id);
$roles = Role::where('title', '!=', 'Super Admin')->get();
return View('admin.user.edit', compact(['data', 'roles']));
}
...
with the above code if I visit http://my.url/users/10/edit I get NotFoundHttpException in Application.php line 901:, yes because there is no id 10 in my record, but with User::find($id); I get normal view without data, since no id 10 in my record.
What I want is show default 404 then redirect to somewhere or return something if record not found with User::findOrFail($id); ? How I can do that ?
Thanks, any help appreciated.
ps: .env APP_DEBUG = true

Answer 1

This does what you asked. No need for exceptions.
public function edit($id)
{
$data = User::find($id);
if ($data == null) {
// User not found, show 404 or whatever you want to do
// example:
return View('admin.user.notFound', [], 404);
} else {
$roles = Role::where('title', '!=', 'Super Admin')->get();
return View('admin.user.edit', compact(['data', 'roles']));
}
}
Your exception handler is not necessary as it is. Regarding Illuminate\Database\Eloquent\ModelNotFoundException:
If the exception is not caught, a 404 HTTP response is automatically sent back to the user, so it is not necessary to write explicit checks to return 404 responses when using [findOrFail()].
Also, I'm pretty sure you get the exception page instead of 404 now because you're in debug mode.

Answer 2

public function singleUser($id)
{
try {
$user= User::FindOrFail($id);
return response()->json(['user'=>user], 200);
} catch (\Exception $e) {
return response()->json(['message'=>'user not found!'], 404);
}
}

Answer 3

findOrFail() is alike of find() function with one extra ability - to throws the Not Found Exceptions
Sometimes you may wish to throw an exception if a model is not found. This is particularly useful in routes or controllers. The findOrFail and firstOrFail methods will retrieve the first result of the query; however, if no result is found, a Illuminate\Database\Eloquent\ModelNotFoundException will be thrown:
$model = App\Flight::findOrFail(1);
$model = App\Flight::where('legs', '>', 100)->firstOrFail();
If the exception is not caught, a 404 HTTP response is automatically sent back to the user. It is not necessary to write explicit checks to return 404 responses when using these methods:
Route::get('/api/flights/{id}', function ($id) {
return App\Flight::findOrFail($id);
});
Its not recommended but If still you want to handle this exception globally, following are the changes as per your handle.php
/**
* Render an exception into an HTTP response.
*
* #param \Illuminate\Http\Request $request
* #param \Exception $exception
* #return \Illuminate\Http\Response
*/
public function render($request, Exception $exception)
{
if ($exception instanceof \Illuminate\Database\Eloquent\ModelNotFoundException) {
//redirect to errors.custom view page
return response()->view('errors.custom', [], 404);
}
return parent::render($request, $exception);
}

Answer 4

Late addition to above topic: If you want to handle the exception for an API backend and you don't want to make the check for an empty result in each method and return a 400 Bad request error individually like this...
public function open($ingredient_id){
$ingredient = Ingredient::find($ingredient_id);
if(!$ingredient){
return response()->json(['error' => 1, 'message' => 'Unable to find Ingredient with ID '. $ingredient_id], 400);
}
return $ingredient;
}
Instead use findOrFailand catch exception in app/Exceptions/Handler.php.
public function render($request, Exception $exception)
{
if ($exception instanceof \Illuminate\Database\Eloquent\ModelNotFoundException) {
return response()->json(['error'=>1,'message'=> 'ModelNotFoundException handled for API' ], 400);
}
return parent::render($request, $exception);
}
This will then look like this in your Controllers:
public function open($ingredient_id){
return Ingredient::findOrFail($ingredient_id);
}
which is much cleaner. Consider that you have plenty of Models and plenty of Controllers.