Using Interface without using argument in function on laravel

Question

i developing a laravel 5 package and now i want to use: interface Gate, on laravel.

in order to use interfaces, i should do by this way:

     public function boot(GateContract $gate)
      {
...code...

$gate->define($permission->name)

    ...Codes...
       }

but i want to know is there another way use GateContract $gate without using argument of function or with using extend from class.

i mean, can i use like this way:

$gate = new Gate;

or any thing else,

Margus Pala · Accepted Answer · 2016-02-14 12:44:23Z

1

Try something like that

$gate = $this->app->make('Gate');

There is more information about the Laravel Service container which instantiates objects over here https://laravel.com/docs/5.2/container

answered Feb 14, 2016 at 12:44

Margus Pala

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