I try to build a path for a model on laravel
I created a function in my model:
public function path()
{
return App\Helper\GeneralController::getURL($this);
}
with dd(App\Helper\GeneralController::getURL($this)) test I got the right answer. (output is a URL)
but in view with the call: $article->path I get this error:
App\Article:: path must return a relationship instance.
What is wrong?
You need to call it:
$article->path()
When you do $article->path, you're trying to use Eloquent relationship which you don't have.
I know this has already been answered and accepted. However, if the OP did want to use a property accessor rather than a method use the "get{property name}Attribute" syntax of Laravel to create a custom attribute.
Here is what it would look like for this specific case:
public function getPathAttribute()
{
return App\Helper\GeneralController::getURL($this);
}
using this approach "path" can now be called as an attribute and will not be resolved to a relationship using the syntax:
$article->path;
You're calling a relationship.
$article->path
To call the method, use '()', like so,
$article->path()
I faced that error when I forgot to write return before relation in the model!
check it out now!
path() is method not object element you need to call as method
$article->path();
Laravel 9 introduced a new way to define accessors/mutators within a model using Illuminate\Database\Eloquent\Casts\Attribute.
https://laravel.com/docs/9.x/eloquent-mutators#defining-an-accessor
public function path(): Attribute
{
return new Attribute(fn () => GeneralController::getURL($this));
}
For future visitors from Google, all the other answers can be applicable in certain scenarios, but you might want to also look if your method access modifier, if your method is protected and you try to call it you will be welcome with this error. You need change your method to public.
Related
When using custom keys Laravel forces us with scoping, for example, I have a route to getting a country and a post
api/countries/{country:slug}/posts/{post:slug}
but I can't get that using slug key because it doesn't have a relation with country, and in this case, I want to handle scope myself and I don't need implicitly scope binding, but I get an error (Call to undefined method App\Country::posts() ).
so because of that I cant using this Laravel feature. is there a way to turn the implicitly scope binding off?
If the posts are not related to the countries, it may not make sense to nest them in the URI?
But, nonetheless, to answer your question, you need to do one of two things:
Instead of setting {country:slug}, just use {country} and then override getKeyRouteName() function on your Country and Post models.
Alternatively, especially if you want to use the ID elsewhere, use explicit model binding.
To use a slug without custom keys in the routes file
class Post
{
[...]
public function getRouteKeyName()
{
return 'slug';
}
}
To use explicit route model binding
Add the following to the boot() method of your RouteServiceProvider:
public function boot()
{
parent::boot();
Route::bind('post', function ($value) {
return App\Post::where('slug', $value)->firstOrFail();
});
}
I'm new to OOP and Yii2. I have a function in Model:
public function getDatRev() {
if ($this->rev) {
return $this->rev;
} else {
return $this->datum;
}
}
in the View until now I have used it like this:
$model->datRev;
and it would return the correct value. Now I don't know what has changed, maybe I was also updated the framework, but the old construct doesn't work anymore, and in order to make it work I have to change it to:
$model->getDatRev();
Can you please explain to me why that is?
When you try get property the Yii2 calls magic method __get (). Return value is depend from implementation of this method in parent class. Yii2 can check if this property exist in some container, or if exist getter of this property.
In your case seems like you don't call parent's method __get(). This may have happened because you override __get() method or initialized this property.
Your class needs to extend yii\base\Object (directly or not) in order to use short property syntax ($model->abc instead of $model->getAbc()). Magic method __get() #Timur mentioned is defined there and further extended in yii\base\Component class.
I am new to Laravel so am uncertain of the notation. The Laravel documentation shows you can pass a parameter this way:
Route::get('user/{id}', function ($id) {
return 'User '.$id;
});
Which is very intuitive. Here is some existing coding in a project I'm undertaking, found in routes.php:
Route::get('geolocate', 'Api\CountriesController#geolocate');
# which of course will call geolocate() in that class
And here is the code I want to pass a variable to:
Route::get('feed/{identifier}', 'Api\FeedController#feed');
The question being, how do I pass $identifier to the class as:
feed($identifier)
Thanks!
Also one further sequitir question from this, how would I notate {identifier} so that it is optional, i.e. simply /feed/ would match this route?
You should first create a link which looks like:
/feed/123
Then, in your controller the method would look like this:
feed($identifier)
{
// Do something with $identifier
}
Laravel is smart enough to map router parameters to controller method arguments, which is nice!
Alternatively, you could use the Request object to return the identifier value like this:
feed()
{
Request::get('identifier');
}
Both methods have their merits, I'd personally use the first example for grabbing one or two router parameters and the second example for when I need to do more complicated things.
I am working with laravel 4.2 and have table in db with property is_active.
When I try to access this model property:
$model->is_active
I am getting following error:
Relationship method must return an object of type Illuminate\Database\Eloquent\Relations\Relation
So question is how to access this property?
Please do not recommend to rename this field in the database if possible because this is already existing database in production.
Here is my model class:
class Position extends \Eloquent {
protected $table = "hr_positions";
protected $fillable = ['slug', 'info_small', 'info_full', 'is_active', 'start_date', 'end_date', 'tags', 'user_create_id', 'user_update_id'];
use \MyApp\Core\StartEndDateTrait;
public function postulations(){
return $this->hasMany('Postulation', 'position_id', 'id');
}
}
Latest notice:
All this error ocurrs on a page where I am creating my entity. In the controller before forwarding to the page I am doing:
$position = new \Position();
and then, for example, following code produce error as well:
dd(($position->getAttribute('is_active')));
but if I replace $position = new \Position(); with
$position = \Position::first();
error is gone?
What is going on here?????
Laravel does a lot of magic behind the scenes, as in, calls a lot of php magic methods.
If a called property is not defined, __call is invoked which in Eloquent calls getAttribute().
Steps taken by getAttribute($key) are
Is there a database field by this key? If so, return it.
Is there a loaded relationship by this key? If so, return it.
Is there a camelCase method of this key? If so, return it. (is_active looks for isActive method)
Returns null.
The only time that exception is thrown is in step 3.
When you create a new instance, eloquent has no idea what kind of fields it has, so if you have a method by the same name, it will always throw a relation error, this seems to be the case in both Laravel4 and Laravel5.
How to avoid it? Use the getAttributeValue($key) method. It has no relation checks and returns null by default.
Alternatively you can also add a get mutator for your field.
I have found a hack for this. Still not ideal but at least I have some solution. Better any than none.
So This code produce problem:
$position = new \Position();
if($position->is_active){
//
}
and this one works fine, this is solution even hacky but solution:
$position = new \Position(['is_active' => 0]);
if($position->is_active){
//
}
I will wait if someone give better, cleaner solution. If no one comes in next few days I will accept mine.
I read a lot about Laravel4 Route-model binding (L4 docs, tutorials, etc.) but still exceptions (i.e. the model is not found) don't work for me
These are my basic files
routes.php:
Route::model('game', 'Game', function(){
// override default 404 behavior if model not found, see Laravel docs
return Redirect::to('/games');
});
...
Route::get('/games/edit/{game}', 'GamesController#edit');
GamesController.php
class GamesController extends BaseController {
...
public function edit(Game $game){
return View::make('/games/edit', compact('game'));
}
}
Pretty straight, but I get this error: Argument 1 passed to GamesController::edit() must be an instance of Game, instance of Illuminate\Http\RedirectResponse given
If I type http://mysite.dev/games/edit/1 all is fine (model with ID = 1 exists)
If I type http://mysite.dev/games/edit/12345 (no model with that ID) the ugly error above is triggered instead of the redirect I specified
I also looked at this (the bottom part where a Redirect closure is suggested: that is just what I am doing!) but no way to make it work: laravel 4 handle not found in Route::model
What's wrong with it? Please any help?
Thanks in advance
In Route::model you declare which variable will be a model instance, you shouldn't use it to do a redirection that way. Instead of that, specify that $game is of type Game and then work with your routes:
Route::model('game', 'Game');
...
Route::get('/games/edit/{game}', 'GamesController#edit');
Then if you access to /games/edit/3 GamesController::edit will receive an instance of Game class whose id=3
I ended up by setting a general "Not Found" error catcher, like this:
// routes.php
App::error(function(Symfony\Component\HttpKernel\Exception\NotFoundHttpException $e) {
return Response::make('Not Found', 404);
});
...
Route::model('game', 'Game');
...
Route::get('/games/edit/{game}', 'GamesController#edit');
What I understand is that if I want a custom redirect and not a general 404 page (i.e. take the user to games' list if model not found), I CAN'T use the route-model-binding
In other words, I have to use Route::get('/games/edit/{id}', 'GamesController#edit'); and then do my application logic inside the 'edit' method:
public function edit($id){
$game = Game::findOrFail($id);
// if fails then redirect to custom page, else go on saving
}
I'm very new to Laravel, but as far as I can see this has nothing to do with the closure, but with the use of "Redirect::to" inside that closure. Using "App::abort( 404 );" works.