Question

我正在使用codeigniter开发一个项目。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创建单独的条件，但我需要一个优化的方法，因为我稍后会有超过25个属性，所以25个条件看起来很合适。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，是否有任何优化方法可以做到这一点？像

这样的东西

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

还是其他任何方式？提前致谢

Answer 1

有25个功能，您还可以查看25列。如果功能的数量经常变化，为什么不能拥有一对listing_id和feature列？这样，只需要将存在的功能插入到数据库中。

您的WHERE查询可以

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

其中必须存在所有指定的条件。当然，由于您现在要将hotel_features中的多行连接到listings中的单行，您应该汇总来自hotel_features的行：

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

Answer 2

$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";   

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

    $counter    =   0;
    foreach($room_features_array as $key => $value){

        if($value == $room_features_array[$ids[$counter]]){
            $this->db->where("hotel_features.$value",1);    
        }
        $counter++;
    }
}

Answer 3

使用$ this-＆gt; db-＆gt; where_in（）;用于检查多个字段包含单个值，反之亦然

在你的情况下，

$condn_arr = array('hotel_features.extra_beds','hotel_features.satellite_tv','hotel_features.airconditioning'); 
$this->db->where(1,$condn_arr);

优化在codeigniter中进行查询的方式

3 个答案: