Php Laravel-如何质疑雄辩的关系?
我在Laravel-5.8中有这个模型:Php Laravel-如何质疑雄辩的关系?,php,laravel,laravel-5.8,Php,Laravel,Laravel 5.8,我在Laravel-5.8中有这个模型: class Employee extends Model { public $timestamps = false; protected $table = 'employees'; protected $primaryKey = 'id'; protected $fillable = [ 'id', 'first_name',
class Employee extends Model
{
public $timestamps = false;
protected $table = 'employees';
protected $primaryKey = 'id';
protected $fillable = [
'id',
'first_name',
'last_name',
'hr_status',
'employee_type_id',
];
public function employeetype()
{
return $this->belongsTo('App\Models\Hr\EmployeeType','employee_type_id','id');
}
}
class EmployeeType extends Model
{
public $timestamps = false;
protected $table = 'employee_types';
protected $primaryKey = 'id';
protected $fillable = [
'type_name',
'is_active',
];
}
然后我在Employee controller函数中进行了以下查询:
$unsubmitted = Employee::where('hr_status', 0)->get();
如何将员工类型中的where is_active=1包含到员工的查询中:
$unsubmitted = Employee::where('hr_status', 0)->get();
您正在查找whereHas方法,查询是否存在关系:
Employee::where('hr_status', 0)
->whereHas('employeetype', function ($q) {
$q->where('is_active', 1);
})->get();
where has
因此,您希望所有拥有hr\u status=0
且与任何拥有is\u active=1
的employeetype有关系的员工都有关系?@lagbox-是的,您是对的。注意:您编写了Laravel 5.8-这是EOL(生命终结)。您应该考虑更新到支持的版本。有关详细信息,请参阅。