我正在Codeigniter中从事一个项目.我在基于多个条件进行查询时遇到问题.最简单的方法是为每个属性设置单独的条件,但是我需要一种优化的方法,因为以后我将拥有25个以上的属性,因此25个条件看起来很奇怪.
这是示例代码

$this->db->select('*');
$this->db->from('listings');
$this->db->join('hotel_features','listings.id = hotel_features.listing_id');
$this->db->where('listings.country_code',$country_t);
$this->db->like('listings.city',$city_t);


    if($room_features_ids != '')
    {
        $room_features_array[0] = "extra_beds";
        $room_features_array[1] = "satellite_tv";
        $room_features_array[2] = "airconditioning";
        $room_features_array[3] = "cable_tv_service";
        $room_features_array[4] = "bathroom";
        $room_features_array[5] = "phone";
        $room_features_array[6] = "wifi";
        $room_features_array[7] = "kitchen";
        $room_features_array[8] = "desk";
        $room_features_array[9] = "refrigerator";   

        $room_features_ids = explode("-",$room_features_ids);

                    // if $room_features_array has 0,1,2 this means first 3 features are available in hotel.
        foreach($room_features_ids as $ids)
        {
            if($room_features_array[$ids] == 'extra_beds')
            {
                $this->db->where('hotel_features.extra_beds',1);    
            }
            else if($room_features_array[$ids] == 'satellite_tv')
            {
                $this->db->where('hotel_features.satellite_tv',1);  
            }
            //and so on.... for all properties
        }
    }

现在我的问题是,有没有优化的方法？
就像是

        foreach($room_features_ids as $ids)
        {
            $this->db->where('hotel_features.'.$room_features_array[$ids],1);   
        }

还是其他方式？提前致谢

解决方法:

具有25个功能,您还将查看25列.如果功能部件的数量经常更改,为什么不使用一对listing_id和功能部件列呢？这样,只需将存在的功能插入数据库即可.

您的WHERE查询可以是

foreach($room_features_ids as $ids)
    {
        $this->db->where('hotel_features.feature', $room_features_array[$ids]);
    }

必须满足所有指定条件的地方.当然,由于您现在要将hotel_features的多个行连接到列表中的单个行,因此您应该汇总hotel_features的行：

$this->db->group_by('listings.id);