雄辩的查询范围返回生成器,而不是使用phpunit时的模型 [英] Eloquent Query Scope return Builder instead of Model when using phpunit
问题描述
我有以下代码
$user = User::findByAccountCode($transaction->account_code);
当我在phpunit上执行此代码时,它将返回Illuminate \ Database \ Eloquent \ Builder的实例,而不是用户模型.
When I execute this code on phpunit it returns an instance of Illuminate\Database\Eloquent\Builder instead of User Model.
这是findByAccountCode的代码
Here is the code for findByAccountCode
public function scopeFindByAccountCode($query,$account_code){
return $query->where('account_code', $account_code)->first();
}
我的应用程序出现以下错误
I get the following error on my Application
ErrorException:参数1传递给 aunicaj \ Libraries \ MarkupRepository :: user()必须是以下对象的实例 aunicaj \ Models \ User,Illuminate \ Database \ Eloquent \ Builder的实例 给
ErrorException: Argument 1 passed to aunicaj\Libraries\MarkupRepository::user() must be an instance of aunicaj\Models\User, instance of Illuminate\Database\Eloquent\Builder given
当我使用浏览器时,它可以正常运行,但不能在phpunit上运行.谢谢
When I use the browser its working fine but not on phpunit. Thank you
推荐答案
我解决了问题,结果发现我测试中的工厂方法正在使用
I got the problem solve turns out that my factory method in my test is using
factory(User::class)->make()
应该是
factory(User::class)->create()
这篇关于雄辩的查询范围返回生成器,而不是使用phpunit时的模型的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!