I want to pass current user's id into a column by default. I tried giving it in the migration but didn't work, this code did work when I pass an integer but it gives an error when I try to set it to Auth::id()
Code I've tried (in the model file)
protected $attributes = [
'employee_id' => Auth::id(),
];
Error I get :
Constant expression contains invalid operations
It does work when I give it a hard coded string or integer value. But I need to give it the current user's id.
Not sure if it's really a good idea, but you can add this in your Model
protected static function booted()
{
static::creating(function ($model) {
$model->employee_id = Auth::id();
});
}
Check the docs for the complete list of event.
https://laravel.com/docs/9.x/eloquent#events-using-closures
Related
I am trying to update a row in the pages table.
The slug must be unique in the pages table on the slug and app_id field combined.
i.e. there can be multiple slugs entitled 'this-is-my-slug' but they must have unique app_id.
Therefore I have found that formula for the unique rule is:
unique:table,column,except,idColumn,extraColumn,extraColumnValue
I have an update method and getValidationRules method.
public function update($resource,$id,$request){
$app_id=22;
$request->validate(
$this->getValidationRules($id,$app_id)
);
// ...store
}
When I test for just a unique slug the following works:
public function getValidationRules($id,$app_id){
return [
'title'=> 'required',
'slug'=> 'required|unique:pages,slug,'.$id
];
}
However, when I try and add the app_id into the validation rules it returns server error.
public function getValidationRules($id,$app_id){
return [
'title'=> 'required',
'slug'=> 'required|unique:pages,slug,'.$id.',app_id,'.$app_id
];
}
I have also tried to use the Rule facade, but that also returns server error. Infact I can't even get that working for just the ignore id!
public function getValidationRules($id,$app_id){
return [
'title'=> 'required',
'slug'=> [Rule::unique('pages','slug')->where('app_id',$app_id)->ignore($id)]
];
}
Any help is much appreciated :)
Thanks for the respsonses. It turned out a couple of things were wrong.
Firstly if you want to use the Rule facade for the validation rules, make sure you've included it:
use Illuminate\Validation\Rule;
The other method for defining the validation rule seems to be limited to the following pattern:
unique:table,column,except,idColumn
The blog post that I read that showed you could add additional columns was for laravel 7, so i guess that is no longer the case for laravel 9.
Thanks for your responses and help in the chat!
I recommend you to add your own custom rule.
First run artisan make:rule SlugWithUniqueAppIdRule
This will create new file/class inside App\Rules called SlugWIthUniqueAppRule.php.
Next inside, lets add your custom rule and message when error occured.
public function passes($attribute, $value)
{
// I assume you use model Page for table pages
$app_id = request()->id;
$pageExists = Page::query()
->where('slug', $slug)
->where('app_id', $app_id)
->exists();
return !$pageExists;
}
public function message()
{
return 'The slug must have unique app id.';
}
Than you can use it inside your validation.
return [
'title'=> 'required|string',
'slug' => new SlugWithUniqueAppIdRule(),
];
You can try it again and adjust this custom rule according to your needs.
Bonus:
I recommend to move your form request into separate class.
Run artisan make:request UpdateSlugAppRequest
And check this newly made file in App\Http\Requests.
This request class by default will consists of 2 public methods : authorize() and rules().
Change authorize to return true, or otherwise this route can not be accessed.
Move your rules array from controller into rules().
public function rules()
{
return [
'title'=> 'required|string',
'slug' => new SlugWithUniqueAppIdRule(),
];
}
To use it inside your controller:
public function update(UpdateSlugAppRequest $request, $resource, $id){
// this will return validated inputs in array format
$validated = $request->validated();
// ...store process , move to a ServiceClass
}
This will make your controller a lot slimmer.
I need to block users from visualizing other users profiles I have the following in my web.php:
Route::get('companyuser/{id}', [CompanyUserController::class, 'show'])
->middleware(['role:companyuser', 'can:show,user']);
I defined a policy
public function show(User $authenticatedUser, $user_model)
{
return $authenticatedUser === $user_model->id ? Response::allow() : Response::deny();
}
and added it to the AuthServiceProvider
protected $policies = [
'App\Model' => 'App\Policies\ModelPolicy',
User::class => CompanyUserPolicy::class,
];
But now the user is blocked from entering his own profile as well. What am I missing? Thanks for the help.
It looks like the policy show() method is comparing an object to an ID value. This will always return false because the $authenticatedUser variable is a User object, while $user_model->id is likely an integer. Using strict type comparison === an int can never be equal to an object:
$authenticatedUser === $user_model->id
Instead the code should probably compare the id of both objects:
$authenticatedUser->id === $user_model->id
Laravel Models also have a built-in method public function is($model): bool that can be used to verify two objects represent the same model (database record). Here is the same code using that method:
$authenticatedUser->is($user_model)
The final solution might look like this:
public function show(User $authenticatedUser, $user_model)
{
return $authenticatedUser->is($user_model) ? Response::allow() : Response::deny();
}
I have this custom function for atempting to login in Laravel 8
protected function attemptLogin(Request $request)
{
$credentials = $this->credentials($request);
$credentials['estado']=1;
return $this->guard()->attempt(
$credentials, $request->filled('remember')
);
}
How I can make to accept the login atempt when $credentials['estado'] also has 2 as value.
Don't know how to make it accept multiple values.
I managed to make the custom function accept the value of 1 but dunno how to make it accept multiple $credentials['estado'] values.
You don't need to change anything in attemptLogin() method, instead you can customize the crededentials() method in LoginController like this:
// login, if user have like a following described data in array
protected function credentials(Request $request)
{
$username = $this->username();
return [
$username => $request->get($username),
'password' => $request->get('password'),
'estado' => [ 1, 2 ], // OR condition
];
}
Answer for comments:
Honestly in my experience I didn't have that case, but if you want to redirect to the another view on failed login (for specific field 'estado'), you can customize the "sendFailedLoginResponse" method, and add some additional if-condition for checking the 'estado'.
As the "sendFailedLoginResponse" method will be called only for getting failed login response instance, then you can check: is that fail comes from 'estado' field actually. Something like this:
protected function sendFailedLoginResponse(Request $request)
{
// custom case, when login failed and estado is 2
if ($request->get('estado') == 2) {
return view('some.specific.view');
}
// laravel by default implementation
else {
throw ValidationException::withMessages([
$this->username() => [trans('auth.failed')],
]);
}
}
Remember, in this case (when we're redirecting the user to some page) we actually not redirecting as for always, but instead we're just returning a view. We do that because I think you don't want to let the users to open that specific view page anytime their want, as you need to let them see that page only for specific case. But when you'll do the actual redirect, then you will let the users to visit that page with some static URL.
Of course, you can do some additional stuff (add something in DB or the Session, and check is the request comes actually from 'estado' fails, but not from any user), but this could be a headeche for you, and in my opinion that will not be a real laravel-specific code.
Anyway, this is the strategy. I don't think, that this is mandatory, but this can be do your work easy and secure.
Note: I've got this deafault implementations from "AuthenticatesUsers" trait (use use Illuminate\Foundation\Auth\AuthenticatesUsers;). In any time you can get some available methods from there and override them in your LoginController, as the LoginController used that as a trait.
I wonder if I should do form validation before retrieving input values or vice versa.
I usually do validation first as I see no benefit in trying to access input values that might not be valid. However, a coworker looked at my code recently and found it strange. Is there any correct order for these steps?
public function createGroups(Request $request)
{
$this->validate($request, [
'courses' => 'required_without:sections',
'sections' => 'required_without:courses',
'group_set_name' => 'required',
'group_number' => 'required|integer|min:1'
]);
$courses = $request->input('courses');
$sections = $request->input('sections');
$group_set_name = $request->input('group_set_name');
$group_number = $request->input('group_number');
Positioning the validation for your controller logic at the beginning of a method is probably the way to go here, as you have required parameters defined. If you receive data that does not fully satisfy the requirements, you produce a validation error back to the user. This follows the productive "Fail Fast" line of thinking: https://en.wikipedia.org/wiki/Fail-fast
It's also important that you're not using any data that hasn't passed your stringent requirements from validation. Data that fails validation should no longer be trusted. Unless there's some other reason you need to be, say, logging any incoming data from the frontend, the order here looks good to me.
I totally agree with #1000Nettles response, to elaborate a little bit more on his/her answer (who should be the accepted one): There isn't any need to continue with your business logic when the data doens't comply with your specifications. Let's say you expected a string of a N characters long, because you defined your database with that limitation (in order to optimize the db desing), will you try to persist it even when it'll throw an exception? Not really.
Besides, Laravel has a particular way to extract validation classes: Form Request. This are injected in controllers. When a call reach the controller it means that already passed the validation, if not, an 422error be returned.
Create a custom request and keep the mess out of your controller, it doesn't even hit your controller function if validation failed and can just grab the data in your controller if validation passed.
php artisan make:request GroupRequest
In app/Http/Requests/GroupRequest.php:
public function authorize()
{
// return true;
return request()->user()-isAdmin; // <-- example, but true if anyone can use this form
}
public function rules()
{
return [
'courses' => ['required_without:sections'],
'sections' => ['required_without:courses'],
'group_set_name' => ['required'],
'group_number' => ['required', 'integer', 'min:1'],
];
}
The best part is you can even manipulate the data in here (GroupRequest.php) after it has been validated:
public function validated()
{
$validated = $this->getValidatorInstance()->validate();
// EXAMPLE: hash password here then just use new hashed password in controller
$validated['password'] = Hash::make($validated['password']);
return $validated;
}
In your controller:
public function createUser(UserRequest $request) // <- in your case 'GroupRequest'
{
$validated = $request->validated(); // <-- already passed validation
$new_user = User::create($validated); // <-- password already hashed in $validated
return view('dashboard.users.show')->with(compact('user'));
}
In your case, if you use my GroupRequest block above, you can return to view in 1 line of code:
public function createGroups(GroupRequest $request)
{
return view('example.groups.show')->with($request->validated()); // <-- already an array
}
In you blade view file, you can then use your variables like {{ $group_set_name }} and {{ $group_number }}
I have a Printer model which has a page_count field..
the user will be able to input the current page_count...
the new page_count must be greater than the existing data in the database... How can I do that?
I had the same issue solved like this, though someone already gave the solution in the comments section.
/**
* #param array $data
* validates and Stores the application data
*
*/
public function sendMoney(Request $request)
{
//get the value to be validated against
$balance = Auth::user()->balance;
$validator = Validator::make($request->all(), [
'send_to_address' => 'required',
'amount_to_send' => 'required|max:'.$balance.'|min:0.01|numeric',
]);
//some logic goes here
}
Depending on your use case you could modify...
Happy Coding
Assuming you have Printer model which contains the page_count column.
You can define a custom validation rule in your AppServiceProvider's boot() method.
public function boot()
{
//your other code
Validator::extend('page_count', function($attribute, $value, $parameters, $validator) {
$page_count = Printer::find(1)->first()->value('page_count'); //replace this with your method of getting page count.
//If it depends on any extra parameter you can pass it as a parameter in the validation rule and extract it here using $parameter variable.
return $value >= $page_count;
});
//your other code
}
Then, you can use it in your validation rule like below
'page_count' => 'required|page_count'
Reference: Laravel Custom Validation