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设置单独的条件,但我需要一种优化的方法,因为稍后我将拥有超过 25 个属性,所以 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')
{
$this->db->where('hotel_features.satellite_tv',1);
}
//and so on.... for all properties
}
}
现在我的问题是,有什么优化方法吗? 有点像
foreach($room_features_ids as $ids)
{
$this->db->where('hotel_features.'.$room_features_array[$ids],1);
}
或者其他方式?提前致谢
最佳答案
有了 25 个特征,您还要查看 25 个列。如果功能的数量经常变化,为什么不使用一对 listing_id
和 feature
列呢?这样,只有存在的特征需要插入到数据库中。
你的WHERE
查询可以是
foreach($room_features_ids as $ids)
{
$this->db->where('hotel_features.feature', $room_features_array[$ids]);
}
所有指定的条件都必须存在。当然,由于您现在将 hotel_features
中的多行连接到 listings
中的单行,因此您应该聚合 hotel_features
中的行:
$this->db->group_by('listings.id);
关于php - 优化在 codeigniter 中进行查询的方式,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/19288079/