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I have a users table and a one-to-zero/one relation with a businesses table (users.user_id => businesses.user_id). On my users table I have a discriminator which tells me if the user is of type business and therefore I need to have details on the businesses table as well.

I want to create my Users with my factory which currently is working and then only create business details where the discriminator points to a business account.

I have three options in my mind:

  1. Create from users factory and then using '->each()' do some checks on the discriminator and create a new business user using a the factory. However I cannot pass to the business factory the user_id that the user was assigned.
  2. First create the users. Then in my Business seeder, retrieve all Users that match a 'business' discriminator. Then for all of these users run a factory that creates the business details. But again, I would have to link somehow the user_id of the already create user with the business factory user_id.
  3. In my business factory, create a new User and retrieve the id, thus making the link between users.user_id and business.user_id. However I am using a random generator for user.user_type so even if I have the businesses table filled it might be for users that have the discriminator as 'personal'.

Is there another way? Can I pass arguments from my Seeder to the factory?

Cristian
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3 Answers3

108

The attributes you pass to the create function will be passed into your model definition callback as the second argument.

In your case you don't even need to access those attributes, since they'll automatically be merged in:

$business = factory(App\Business::class)->create();

factory(App\User::class, 5)->create([
    'business_id' => $business->id,
]);

Adapt this to your needs.

Joseph Silber
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    How could I have missed that. I watched the Laracast lesson and I know remember about setting stuff myself. Of course it was in the documentation as well: http://laravel.com/docs/master/testing#model-factories. Thanks a lot. – Cristian Sep 03 '15 at 14:47
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    @Cristian's link looks outdated now, for 5.4 see https://laravel.com/docs/5.4/database-testing#using-factories & scroll down to 'Persisting Models' – Harry Feb 21 '17 at 16:02
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    what about like this? ```php $student->assignRole('student'); ``` im using spatie/permission – Lloric Mayuga Garcia Oct 07 '17 at 14:31
  • what if i don't want to merge them? what if i want to use custom parameters to model factories? – shotex Oct 08 '17 at 18:17
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    @shotex - If you want custom parameters, why are you using factories? Just create the model yourself: `User::create($attributes)` – Joseph Silber Oct 11 '17 at 15:00
  • @JosephSilber In my case I would want to pass a $user parameter so I can get its type and according to that, get a random value associated with it. – Víctor Oct 11 '18 at 16:12
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    Great, the manual puts it like this: You may override attributes on the model by passing an array to the create method. https://laravel.com/docs/5.8/database-testing#persisting-models – PaulH Mar 12 '19 at 19:32
10

My code for adding polymorphic 'Admin' users was: 

// run model factory
factory(App\Admin::class, 3)->create()->each(function ($admin) {

    $admin->user()->save(

        // solved: https://laravel.com/docs/master/database-testing#using-factories (Overriding attributes)
        factory(App\User::class)->make([
              'userable_id' => $admin->id,
              'userable_type' => App\Admin::class
        ])
    );
});

Hope this helps.

Robin Hood
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4

Send attribute,

factory(App\User::class)->create(['businessId' => $businessId]);

Retrieve it,

$factory->define(App\User::class, function (Faker $faker, $businessInfo) {
    //$businessInfo['businessId']
});
Lakpriya De Silva
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