Phpunit Laravel出纳伪造测试用户订阅
我用拉威尔的收银员结账,这很好。我试图表现良好,并对代码进行测试,但我很难用订阅“伪装”用户 我试过:Phpunit Laravel出纳伪造测试用户订阅,phpunit,laravel-5,stripe-payments,Phpunit,Laravel 5,Stripe Payments,我用拉威尔的收银员结账,这很好。我试图表现良好,并对代码进行测试,但我很难用订阅“伪装”用户 我试过: $user = App\Models\User::create([ 'first_name' => $this->faker->firstName, 'last_name' => $this->faker->lastName, 'email' => $this->faker->email, 'p
$user = App\Models\User::create([
'first_name' => $this->faker->firstName,
'last_name' => $this->faker->lastName,
'email' => $this->faker->email,
'password' => 'password1234',
'stripe_plan' => 'name_of_plan',
'stripe_active' => 1
]);
$this->be($user);
但是如果我随后选中$user->onPlan('name\u of \u plan')
我会得到false
:(
有什么方法可以做到这一点吗?我相信你会理解,在我有测试来备份之前,我并不真的想启动支付系统!检查“stripe\u plan”和“stripe\u active”是否定义为可供用户填充。如果它们没有定义,那么它可能实际上没有在User::create()中设置这些值,这就是测试失败的原因
class User extends Model implements AuthenticatableContract, CanResetPasswordContract, BillableContract {
use Authenticatable, CanResetPassword, Billable;
protected $table = 'users';
protected $fillable = [
'first_name',
'last_name',
'email',
'password',
'stripe_plan',
'stripe_active'
];
}
天才!非常感谢。