I have a resource in Laravel I have called artists with an ArtistsController. I would like to add filters to some of the pages, but not all. I know I can add a filter to all of the functions/views in the resource controller like so:
public function __construct()
{
$this->beforeFilter('auth', array('except' => array()));
}
How do I add the beforeAuth filter to only a certain view/function? I would like a user to be logged in in order to go the "index" view, but I would like a user to be able to go to the "show" pages without necessarily being logged in:
public function index()
{
$artists = Artist::all();
return View::make('artists.index', compact('artists'))
->with('artists', Artist::all())
->with('artists_new', Artist::artists_new());
}
public function show($id)
{
$artist = Artist::find($id);
return View::make('artists.show', compact('artist'))
->with('fans', Fan::all());
}
Is there a way to do this? Thank you.
Not sure if this helps but you could use the only key instead of the except (if I understand your question correctly).
$this->beforeFilter('auth', array('only' => array('login', 'foo', 'bar')));
Although that would still go in the constructor.
Related
So I am trying to get my mutators and accessors to work in Laravel 9, in my Tag model I have the following:
protected function name(): Attribute
{
return Attribute::make(
get: fn ($value) => strtolower($value),
set: fn ($value) => strtolower($value),
);
}
When displaying the name in my blade view however, the name is not being displayed in lower cases ({{ $tag->name }}), also not when saving a new model to the database.
The following does work btw:
public function getNameAttribute($value)
{
return strtolower($value);
}
Also when using public it does not work:
public function name(): Attribute
Just trying to understand what I am doing wrong here?
I am using Laravel version 9.44
I dont know if the question's content is exactly your code. I had a similiar problem, get and set not working. But it worked in other model files.
I just found the solution once again, ya, once again.
If the attribute(column name) has two words, you have to make it together, first word should be lowercase.
short_name
protected function shortName(): Attribute
{
return Attribute::make(
get: fn ($value) => strtolower($value),
set: fn ($value) => strtolower($value),
);
}
Inject the class at top of page like below
use Illuminate\Database\Eloquent\Casts\Attribute;
I've set up Yii2 REST API with custom actions and everything is working just fine. However, what I'm trying to do is return some data from the API which would include database relations set by foreign keys. The relations are there and they are actually working correctly. Here's an example query in one of the controllers:
$result = \app\models\Person::find()->joinWith('fKCountry', true)
->where(..some condition..)->one();
Still in the controller, I can, for example, call something like this:
$result->fKCountry->name
and it would display the appropriate name as the relation is working. So far so good, but as soon as I return the result return $result; which is received from the API clients, the fkCountry is gone and I have no way to access the name mentioned above. The only thing that remains is the value of the foreign key that points to the country table.
I can provide more code and information but I think that's enough to describe the issue. How can I encode the information from the joined data in the return so that the API clients have access to it as well?
Set it up like this
public function actionYourAction() {
return new ActiveDataProvider([
'query' => Person::find()->with('fKCountry'), // and the where() part, etc.
]);
}
Make sure that in your Person model the extraFields function includes fKCountry. If you haven't implemented the extraFields function yet, add it:
public function extraFields() {
return ['fKCountry'];
}
and then when you call the url make sure you add the expand param to tell the action you want to include the fkCountry data. So something like:
/yourcontroller/your-action?expand=fKCountry
I managed to solve the above problem.
Using ActiveDataProvider, I have 3 changes in my code to make it work.
This goes to the controller:
Model::find()
->leftJoin('table_to_join', 'table1.id = table_to_join.table1_id')
->select('table1.*, table_to_join.field_name as field_alias');
In the model, I introduced a new property with the same name as the above alias:
public $field_alias;
Still in the model class, I modified the fields() method:
public function fields()
{
$fields = array_merge(parent::fields(), ['field_alias']);
return $fields;
}
This way my API provides me the data from the joined field.
use with for Eager loading
$result = \app\models\Person::find()->with('fKCountry')
->where(..some condition..)->all();
and then add the attribute 'fkCountry' to fields array
public function fields()
{
$fields= parent::fields();
$fields[]='fkCountry';
return $fields;
}
So $result now will return a json array of person, and each person will have attribute fkCountry:{...}
Laravel 5.1 has just been released, I would like to know how could I tell the AuthController to get the login & register view from a custom directory? the default is: resources/views/auth...
The trait AuthenticateAndRegisterUsers only has this:
trait AuthenticatesAndRegistersUsers
{
use AuthenticatesUsers, RegistersUsers {
AuthenticatesUsers::redirectPath insteadof RegistersUsers;
}
}
The code you're showing there only fills one function: it tells our trait to use the redirectPath from the AuthenticatesUsers trait rather than the one from RegistersUsers.
If you check inside the AuthenticatesUsers trait instead, you will find a getLogin() method. By default, this one is defined as
public function getLogin()
{
return view('auth.login');
}
All you have to do to get another view is then simply overwriting the function in your controller and returning another view. If you for some reason would like to load your views from a directory other than the standard resources/Views, you can do so by calling View::addLocation($path) (you'll find this defined in the Illuminate\View\FileViewFinder implementation of the Illuminate\View\ViewFinderInterface.
Also, please note that changing the auth views directory will do nothing to change the domain or similar. That is dependent on the function name (as per the definition of Route::Controller($uri, $controller, $names=[]). For more details on how routing works, I'd suggest just looking through Illuminate\Routing\Router.
for those who is using laravel 5.2, you only need to override property value of loginView
https://github.com/laravel/framework/blob/5.2/src/Illuminate/Foundation/Auth/AuthenticatesUsers.php
public function showLoginForm()
{
$view = property_exists($this, 'loginView')
? $this->loginView : 'auth.authenticate';
if (view()->exists($view)) {
return view($view);
}
return view('auth.login');
}
so to override the login view path, you only need to do this
class yourUserController {
use AuthenticatesAndRegistersUsers, ThrottlesLogins;
.....
protected $loginView = 'your path';
}
I need to return a different view from a controller depending on the route that is requested.
For example: in my application I have clients, devices and campaigns. All have CRUD's created, but in some cases I want to do something like "view clients, delete his campaign and returning to the clients view" but my campaignsController#delete returns to campaigns by default.
The idea is not rewrite the same controller only changing the route of returning, does Laravel have something to help with this?
Thank you
Laravel will not control the whole flow of your application. If you have a campaign delete router:
Route::delete('campaign/{id}');
and it returns to campaigns
class CampaignController extends Controller {
public function delete($id)
{
$c = Campaign::find($id);
$c->delete();
return Redirect::route('campaigns');
}
}
You will have to trick your route to make it go to wherever you need to, there should be dozens of ways of doing that, this is a very simple one:
Route::delete('campaign/{id}/{returnRoute?}');
class CampaignController extends Controller {
public function delete($id, $returnRoute = null)
{
$returnRoute = $returnRoute ?: 'campaigns';
$c = Campaign::find($id);
$c->delete();
return Redirect::route($returnRoute);
}
}
And create the links in those pages by using the return route option:
link_to_route('campaign.delete', 'Delete Campaign', ['id' => $id, 'returnRoute' => 'clients'])
i have a behavior for my models, the behavior has beforeFind, beforeSave, in methods i override user_id, something like:
...
public functio beforeSave() {
$this->owner->user_id = Yii::app()->user->id
}
I have model User, how can i disable behavior for registration new user?
Saving code:
$user = new User();
$user->id = 1332;
$user->field1 = 'data';
$user->save();
but on save i have null in $user->id (because work behavior).
i tried
$user->disableBehaviors();
$user->detachBehavior();
Without result.
Maybe its not right way? I create behaviors for identify users in system (find only user something, save only with user id...), but that if i have new user with full previegies, i should again detach behaviors?
If condition can be changed in future I just pass it as callback parameter into behavior from model.
This give you a bit more control over the condition. Hence, behavior becomes more reusable - if it is used by several models this condition can be unique for each.
Example below is a bit simplified, but you should get the idea.
Behavior:
class SomeBehavior extends CActiveRecordBehavior
{
public $trigger;
public function beforeSave($event)
{
if(!call_user_func($this->trigger))
return;
// do what you need
$this->owner->user_id = Yii::app()->user->id;
}
}
Model:
class SomeModel extends CActiveRecord
{
public function behaviors()
{
$me=$this;
return array(
'some'=>array(
'class'=>'SomeBehavior',
'trigger'=>function() use($me){
return $me->scenario=='some-scenario';
}
)
);
}
}
Also I use PHP 5.3. So, I use closure for trigger callback.
If your PHP version is less than 5.3 - anything callable can be used instead. Check here http://www.php.net/manual/en/function.is-callable.php
Because of behavior is a method, you can declare your own logic inside.
The model knows about its scenario, so there is no problem to return different arrays for different conditions:)
Hope it be helpful for somebody.
You can check Yii::app()-user->isGuest to determine if the user is logged in or not. or you can just try looking for the null. Like this:
if (!Yii::app()->user->isGuest)
$this->owner->user_id = Yii::app()->user->id;
or
if (null !== Yii::app()->user->id)
$this->owner->user_id = Yii::app()->user->id;