我有一张表,用于存储餐厅的评分。如下图所示。
我试图获得所有这些列的平均值,我在其中取得了成功,但我也希望将所有这些平均值的平均值作为主要平均值。
我尝试了以下查询,但平均评分为 3,这是不准确的。我认为 mysql 正在向我返回最终结果的整数值。
return $this->db->select('((ambience + music + service + value_for_money + cleanliness + sanitary + view)/7) as rating, AVG(ambience) as ambience, AVG(music) as music, AVG(service) as service,AVG(value_for_money) as value_for_money, AVG(cleanliness) as cleanliness, AVG(sanitary) as sanitary, AVG(view) as view' )
->where('restaurant_id',$restaurantId)
->get('restaurant_ratings')
->row_array();
当我运行上述查询时,评分字段的平均值为 3。
实际结果是 3.42。
请帮助我了解我做错了什么以及可以做些什么来获得准确的结果。 谢谢
答案 0 :(得分:3)
只需添加 AVG
即可计算评分:
$query =
'AVG((
ambience +
music +
service +
value_for_money +
cleanliness +
sanitary +
view
)/7) as rating,
AVG(ambience) as ambience,
AVG(music) as music,
AVG(service) as service,
AVG(value_for_money) as value_for_money,
AVG(cleanliness) as cleanliness,
AVG(sanitary) as sanitary,
AVG(view) as view';
return $this->db
->select($query)
->where('restaurant_id',$restaurantId)
->get('restaurant_ratings')
->row_array();