Php Laravel何处必须是真实模型，何处必须是相关模型不起作用

我有三张桌子：

交易模式：

class Deal extends Model
{
    protected $guarded = ['id'];

    public function hotel() {
        return $this->belongsTo('App\Hotel');
    }
}

class Hotel extends Model
{
    public function room(){
        return $this->hasMany('App\Room');
    }

    public function deal(){
        return $this->hasMany('App\Deal');
    }
}

class Room extends Model
{

    public function hotel(){
        return $this->belongsTo('App\Hotel');
    }

}

酒店模式：

class Deal extends Model
{
    protected $guarded = ['id'];

    public function hotel() {
        return $this->belongsTo('App\Hotel');
    }
}

class Hotel extends Model
{
    public function room(){
        return $this->hasMany('App\Room');
    }

    public function deal(){
        return $this->hasMany('App\Deal');
    }
}

class Room extends Model
{

    public function hotel(){
        return $this->belongsTo('App\Hotel');
    }

}

房间型号：

class Deal extends Model
{
    protected $guarded = ['id'];

    public function hotel() {
        return $this->belongsTo('App\Hotel');
    }
}

class Hotel extends Model
{
    public function room(){
        return $this->hasMany('App\Room');
    }

    public function deal(){
        return $this->hasMany('App\Deal');
    }
}

class Room extends Model
{

    public function hotel(){
        return $this->belongsTo('App\Hotel');
    }

}

下面的查询工作正常

return $greatDeals = Deal::whereHas('hotel', function ($query) {
                $query->Where('astatus', 1)->Where('status', 0);
            })->get();

但我想询问“酒店”模型，其中有“房间”模型 但是下面的查询显示错误，这个查询格式正确吗

 return $greatDeals = Deal::whereHas('hotel', function ($query) {
                    $query->whereHas('room', function ($query) {
                        $query->Where('astatus', 1)->Where('status', 0);
                    })->get();
                })->get();

错误：

"SQLSTATE[42S22]: Column not found: 1054 Unknown column 'deals.hotel_id' in 'where clause' (SQL: select * from `hotels` where `deals`.`hotel_id` = `hotels`.`id` and exists (select * from `rooms` where `hotels`.`id` = `rooms`.`hotel_id` and `astatus` = 1 and `status` = 0)) ◀"

删除第一个

get（）

：

或者这样做：

Deal::whereHas('hotel.room', function ($query) {
        $query->where('astatus', 1)->where('status', 0);
    })->get();

---

两者都很好，我选择第一个，因为我需要为酒店添加相同的where条件（$query->where（'astatus'，1）->where（'status'，0）；）。@Alexey Mezenin。你无处不在，：）