Laravel 5 form request validation with ID - show - php

I am wanting to validate a resource controller in Laravel, so that the user can only access clinic.show if they're the owner of said clinic.
I correctly validated this using the following:
public function show($id)
{
if (Clinic::where('id', $id)->where('user_id', Auth::id())->exists()) {
return View::make('clinic.show', ['clinic' => Clinic::where('id', $id)
->first()]);
} else {
abort(401, 'Unauthorized action.');
}
}
However, I believe this is bad practice and I should be using the Form Request feature within Laravel.
I have created a ShowClinicFormRequest.php and added the following code:
public function authorize()
{
$clinicId = $this->route('clinic');
return Clinic::where('id', $clinicId)
->where('user_id', Auth::id())
->exists();
}
And within the show function -
public function show($id, ShowClinicFormRequest $request)
{
return View::make('clinic.show', ['clinic' => Clinic::where('id', $id)->first()]);
}
However, the error that I am getting is:
ReflectionException in RouteDependencyResolverTrait.php line 53: Class
App\Http\Controllers\ShowClinicFormRequest does not exist
It obviously doesn't exist within that directory because it isn't a controller.
Any help would be greatly appreciated.

I believe your form request is located in the App\Http\Requests namespace so you need to import the class or use explicitly:
use App\Http\Requests\ShowClinicFormRequest;
or just
public function show($id, \App\Http\Requests\ShowClinicFormRequest $request)
{}
You might also to take a look at filters or middlewares.

Related

Newbie question about controller and request in Laravel in my case

I am new in Laravel, what I try to achieve is very simple thing, I would like to use FormRequest provided by Laravel to do validation of the request, but I encounter some puzzles (which I am sure is easy things to solve if you are experienced in Laravel).
Here is what I tried:
I have route maps to controller:
Route::put('user/{name}', 'UserController#show');
I can get the name parameter in show function:
class UserController {
public function show($name)
{
// validtion rules to apply
...
}
}
I have validation rules to apply to the request, so I decided to create form request by php artisan make:request ShowRequest, which creates the form request class:
class ShowRequest extends FormRequest {
public function authorize()
{
return true;
}
public function rules()
{
return [
// my validation rules here
];
}
}
Since I have above request class, so I refactored the show function in controller to receive the ShowRequest .
class UserController {
public function show(ShowRequest $request)
{
// now I don't need validtion rules in this function
// but how can I access the 'name' parameter now
...
}
}
I have two questions to ask:
Inside the refactored show function, how can I now access the route parameter name ?
If we forget about the parameter is a name (please don't focus on what to validate for name, imaging it is an object or value to validate in general). How to add custom logic for handling validation error instead of using Laravel default behaviour. I want to inject code like dummy code below:
if (!$validator->pass())
{
//my custom code for handling validation failure
}
Where to put my custom code for handling validation error now? I mean I don't know where to have this logic, in controller? in the request class? how?
You still can add the parameter $name in the show() method of your controller as it's part of the routed url more than the validated form/data. (recommanded)
class UserController {
public function show(ShowRequest $request, $name)
{
//...
}
}
You can also access it from the request object
class UserController {
public function show(ShowRequest $request)
{
$request->input('name');
}
}
As for the error messages (not the exception) you can add the messages() method to your ShowRequest::class
class ShowRequest extends FormRequest
{
/**
* #return array
*/
public function messages()
{
return [
'name.required' => 'The name is required',
'name.numeric' => 'The name must be a number',
//...
];
}
}
If you instead need to validate that the name catched by the route is only composed of letter OR really exists as a field in your DB (like a slug of a post) you need to add some validation in your route declaration.
Setup a route that catches request only if it is composed of letters.
Route::get('/user/{name}', 'Controller#show')->where(['name' => '[a-z]+']);
Setup a route that catches request only if the "name" exists in DB:
User.php
Class User //..
{
/**
* Get the route key for the model.
*
* #return string
*/
public function getRouteKeyName()
{
return 'name';
}
}
web.php
//
Route::get('/user/{user:name}', 'Controller#show');
And adapt your controller to take a user directly
class UserController {
public function show(ShowRequest $request, User $user)
{
//...
}
}
You can access the values of the Form Request using this
$validated = $request->validated();
The $validated will have all the values which had been validated by the FormRequest.
To answer your second question, if you want to throw custom validation, you can always use the following
throw ValidationException::withMessages(['name' => 'Something is wrong']);

How is Model::find($id) secure in Laravel?

I have an app where I create entries based on who is signed in. If I use the find($id) method it returns json response. The function is like this:
public function edit($id)
{
$chore = Chore::find($id);
return response()->json($chore);
}
Now if I where to edit the id value I might be able to access other user's data which isn't secure at all.
So I added and extra column user_id that checks who is signed in:
public function edit($id)
{
$chore = Chore::find($id)
->where('user_id', Auth::id());
return response()->json($chore);
}
But of course laravel can't make it easy so it doesn't work. Adding ->get() returns an array instead of a json response.
First of all how is find($id) ever secure in any app that uses authentication and secondly how do I add another condition under the find($id) clause? I need data returned in JSON otherwise I will need to rewrite all my front-end which isn't ideal at this point.
I also tried:
public function edit($id)
{
$chore = Chore::where('id', $id)
->where('user_id', Auth::id());
return response()->json($chore);
}
but no luck
You just need to call the where method before the find method
$chore = Chore::where('user_id', Auth::id())
->find($id);
As an alternative, If you've set up relationships properly on user model
// In User.php
public function chores()
{
return $this->hasMany(Chore::class, 'user_id', 'id');
}
then you could also do
$chore = Auth::user()->chores()->find($id);
While this seems like extra work, it's more convenient and easier to maintain.
If Chore is in a one-to-one relationship with your User model, then you can create a relationship in your User.php model.
public function chore() {
return $this->hasOne(Chore::class);
}
Then in your controller, you could simply call auth()->user()->chore.
The Eloquent's find() method finds only one record by the primary key. If you use additional validation it's perfectly safe. You could use route model binding to simplify the code.
web.php
Route::get('/chore/{chore}/', 'ChoreController#edit');
Then in your controller
public function edit(Chore $chore)
{
if (! $chore->user_id === auth()->id()) {
// throw error or redirect, or whetever
}
return response()->json($chore);
}
To simplify the controller a little bit more, you could create a form request and inject it into controller's method as a regular request. (https://laravel.com/docs/7.x/validation#authorizing-form-requests)
Then you could move the validation into your form request. It should look something like this:
public function authorize() {
return $this->route('chore')->user_id === $this->user()->id
}

is it better to create separate request class for each new method in controller, or edit existing request class in laravel or any better idea?

Is it better to create separate request class for each new method in controller or edit existing request class in laravel or any better idea ?
example
class fooBarController {
public function a(fooBarARequest $r) {
}
public function b(fooBarBrequest $r) {
}
public function c(fooBarCDRequest $r) {
}
public function d(fooBarCDRequest $r) {
}
}
Using extra request classes allows you to define validation rules which your request is checked against before it reaches your controller. You can also handle authorization in the request class. An example would be:
class UpdateAccountEmail extends FormRequest
{
public function authorize()
{
return true; // authorization is handled on route/middleware level
}
public function rules()
{
return [
'new_email' => 'required|email|confirmed',
'new_email_confirmation' => 'required',
];
}
}
So, to sum it up: it does not make sense to use a custom request class for requests which do not have payload that needs to be validated. This means, for a normal GET request we most likely (of course there are exceptions) want to use the normal Request class provided by laravel. A controller like this would be quite normal:
class AccountController
{
public function show(Request $request)
{
return view('account.show', ['user' => $request->user()]);
}
public function edit()
{
return view('account.edit', ['user' => \Auth::user()]);
}
public function updateEmail(UpdateAccountEmail $request)
{
$user = $request->user();
$user->email = $request->input('new_email');
$user->save();
return redirect()->route('account.index');
}
public function logins(Request $request)
{
$logins = $request->user()->logins()
->when($request->get('filter_from'), function ($query, $from) {
$query->where('created_at', '>=', $from);
})
->when($request->get('filter_until'), function ($query, $until) {
$query->where('created_at', '<=', $until);
})
->get();
return view('account.logins', ['logins' => $logins]);
}
}
As you can see, for the GET request that is handled by logins(Request $request), we do not use a custom request class because we don't need to validate anything (well, we could validate the filter parameters, but for simplicity we don't).
The example above also shows different methods of retrieving the current user. Even for that you don't need a request at all.
This is no actual production code, just something off the top of my head...

Pass in an id for Authorization Laravel

In my user controller I have an posts function, which gives access to a sub-resource of users. This is accessed through the /users/{id}/posts endpoint.
I want the pass the $id from the request URL into a UserPolicy method:
public function resource($user, $id)
{
return $user->id === $id;
}
My UserController method:
public function posts(Request $request, $id)
{
$this->authorize('resource', $id);
return response()->json(['events' => []], 200);
}
Is there anyway to do this? I notice that Policy methods seem to ignore anything that isn't an object.
Edit:
I am currently using a helper method for this authorization but would like to move it to my Policy to keep all rules together:
public function authorizeResource($id)
{
if ((int)$id !== (int)$this->auth->user()->id) {
throw new \Exception;
}
}
Laravel needs to know which policy class to use. For that you need to specify the model, in this case passing an array with an instance of user first and then the $id. Laravel uses the spread operator and will inject the $id as a parameter on your callback function.
//UserController.php
public function posts(Request $request, $id)
{
$this->authorize('resource', [User::class, $id]);
return response()->json(['events' => []], 200);
}

Not working laravel Policy for my UsersUpdate

I have
(1/1) HttpException
This action is unauthorized.
I think all should work fine and I have done it right but maybe not.
My controller method:
public function update(Request $request, Users $uzytkownik)
{
$this->authorize('update', $uzytkownik);
return 1;
}
UsersPolicy that is in App\Policies\:
<?php
namespace App\Policies;
use App\Models\Users;
use Illuminate\Auth\Access\HandlesAuthorization;
class UsersPolicy
{
use HandlesAuthorization;
public function update(Users $user)
{
return true;
// return $user->login === auth()->login;
}
}
And in AuthServiceProvider:
protected $policies = [
'App\Model' => 'App\Policies\ModelPolicy',
'App\Models\Users' => 'App\Policies\UsersPolicy',
];
My Users model lays in App\Models\
When I cut $this->authorize('update', $uzytkownik); this line from controller everything works fine and I see '1', when I add it again HttpException.
What do I have wrong here? Thinking and Thinking, looking, I don't see anything bad here.
please make sure that your route is under auth middlware like this :
Route::group(['middleware' => 'auth'], function () {
// ur update route here
});
or in ur controller constructor like this :
public function __construct()
{
$this->middleware('auth');
}
and also like #Laerte said your update policy method should have another parameter of type user which is the user you want to edit, like this :
public function update(Users $userLoggedIn, Users $uzytkownik)
{
return true;
}
In your Policy, you have to add two parameters: The first one is the user logged in, and the second is the actual parameter. Try this in the Policy:
public function update(Users $userLoggedIn, $user)
{
return true;
}

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