Dynamic model binding in trait function

jthinam 0
Viewed 24

I have multiple Policy classes.

and these policies' update, delete, restore functions have the same logic evaluation which is to check if the authenticated user owns the resource.

For example, I have a Post and a Comment model.

Then for PostPolicy and CommentPolicy, both of their update, delete, restore functions will all have:

public function update(User $user, Post $post)
{
    return $user->id == $post->user_id;
}

public function delete(User $user, Post $post)
{
    return $user->id == $post->user_id;
}

public function restore(User $user, Post $post)
{
    return $user->id == $post->user_id;
}

// Also the same with CommentPolicy

With that, I might as well have a trait like this:

trait AuthorizableTrait
{
    public function authorize(User $user, Resource $resource)
    {
        return $user->id == $resource->user_id;
    }

}

So, my question is, is it possible to inject a dynamic instance of the current model inside the trait, for example, Post and Comment models now will become Resource? if so, how?

traits php laravel
3 Answers

There's a couple other ways to handle this too. In your trait, you can simply replace Resource $resource with Model $model. As long as your Post.php, Comment.php and Resource.php Models have are class ... extends Model { ... }, then it'll accept it:

<?php 
namespace App\Models\Traits;

use Illuminate\Database\Eloquent\Model;

trait AuthorizableTrait {
  public function authorize(User $user, Model $model) {
    return $user->id == $model->user_id;
  }
}

If that is too "loose", and you don't want this available for all Models (i.e. in the possible case that not all Models have a user_id column), then you can use Union Types (assuming you are on a compatible PHP version):

<?php 
namespace App\Models\Traits;

use App\Models\Post;
use App\Models\Comment;
use App\Models\Resource;

trait AuthorizableTrait {
  public function authorize(User $user, Post|Comment|Resource $model) {
    return $user->id == $model->user_id;
  }
}
Tim Lewis0

If you don't specify the type of the authorize function second parameter, I think it will be all right.

If you define the following function, without the Resource type:

public function authorize(User $user, $resource)

You will be able to send everything you want in $resource (a post, a comment, etc.) and then the check will work... unless you always have a user_id attribute.

Bastien0

UPDATE:

I decided to use PHPStan.

And the neon script requires that all function's parameter and return type must be specified.

Thus, I ended up specifying Model as $resource's type.

public function authorize(User $user, Model $resource)

However, for some reason, in your use-case, Model is too generic. Then you can do something like what @TimLewis did.

jthinam0
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