Laravel: Set string as name for model

Question

I´m working with polymorphic in Laravel 5.2. I want to set for the name of model string. There si code:

switch ($passport->element_type) {
            case 'Window':      
                $windows = Window::find($passport->element_id);
                break;
            case 'Floor':    
                $floor = Floor::find($passport->element_id);                    
                break;
            case 'Wall':
                $wall = Wall::find($passport->element_id);
            default:
                break;
        }
    };

You can see that variable "$passport->element_type" gives me the name of my model. I don´t want to do it with switch-case. Is it possible to do something like:

$passport->element_type::find($passport->element_id);

or how can I use the variable (element_type) as the name for model?

adamwathan.me/2015/09/03/pushing-polymorphism-to-the-database Adam Wathan describes the solution to this well. I don't have time to make it into an answer so i'll just drop the link here. — Ben Swinburne
– Ben Swinburne, Commented Sep 22, 2016 at 18:08

Adiasz · Accepted Answer · 2016-09-22 17:43:06Z

1

You need to set proper Model relations:

class Passport extends Model
{
    /**
     * Get all of the owning likeable models.
     */
    public function pasportable()
    {
        return $this->morphTo();
    }
}

class Window extends Model
{
    public function pasport()
    {
        return $this->morphOne('App\Pasport', 'pasportable');
    }
}

class Floor extends Model
{
    public function pasport()
    {
        return $this->morphOne('App\Floor', 'pasportable');
    }
}

class Wall extends Model
{
    public function pasport()
    {
        return $this->morphOne('App\Wall', 'pasportable');
    }
}

and next you can access to morph'ed object by:

$passport->element();
// or

Passport::find($id)->element();

and it should return Model of Floor, Wall or Window depending on element_type

answered Sep 22, 2016 at 17:43

Adiasz

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Samsquanch · Accepted Answer · 2016-09-22 17:27:24Z

0

Have you tried it?

If

$passport->element_type::find($passport->element_id);

doesn't work, I know that

$type = new $passport->element_type;
$type::find($passport->element_id);

will.

edited Sep 22, 2016 at 17:27

answered Sep 22, 2016 at 17:08

Samsquanch

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8 Comments

jakub Over a year ago

No it doesn´t work. dd($type) return me "Window". But dd($type::find($passport->element_id)) throws error .. Class 'Window' not found.

Samsquanch Over a year ago

Class 'Window' not found -- it's trying to work but it's not finding your model.

jakub Over a year ago

Yes, but dd(Window::find($passport->element_id)) works

jakub Over a year ago

don´t know why but already $type = new $passport->element_type; throws error Class "Window" not found. I think that problem is that it is looking for class 'Window' with apostrophes and it should only looking for class Window without apostrophes.

Samsquanch Over a year ago

Try it with this: $type= "Window";, see if it still throws that error.

|

jakub · Accepted Answer · 2016-09-23 11:29:22Z

0

Thanks guys for help. Finally works, but I needed to add path:

$type = App::make('\\App\\'.$passport->element_type);

answered Sep 23, 2016 at 11:29

jakub

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