本文介绍了如何使用Codeigniter查询获得每种产品的平均评级?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我有一个评分表,其中列出了用户给定的每种产品的评分,我正在检索所有评分记录,但同时我想获得基于每种产品的平均评分,但我无法获得输出量
I have rating table which have rating for every product given by user, I am retrieving all rating records, but at the same time I want to get avg rating based on per product but I am unable to get ouptput
查询:
$this->db->select('ratings.*');
$this->db->select('select AVG(ratings) as avg_rating from ratings group by product_id');
$this->db->from('ratings');
$this->db->join('users','users.id=ratings.user_id');
$this->db->get()->result();
评分表
id user_id product_id rating
1 4 1 4
2 5 2 4
3 6 1 2
4 7 4 4
预期输出:
id user_id product_id rating avg rating
1 4 1 4 3
2 5 2 4 4
3 6 1 2 3
4 7 4 4 4
推荐答案
使用左连接和select作为平均值,从表 ratings 中获取数据.
get data from table ratings, using a left join with select for the average.
的 join()函数Codeigniter允许您编写选择部分而不是表名,但是需要将其放在括号中:
the join() function of Codeigniter allows you to to write a select part instead of the table name, but you need to place it into parenthesis:
$this->db->select('t1.*, t2.avg_rating, t3.*');
$this->db->from('ratings t1');
$this->db->join('
(select product_id, avg(rating) as avg_rating
from ratings
group by product_id) t2','t2.product_id=t1.product_id','left'
);
$this->db->join('users t3','t3.id=t1.user_id','left');
$this->group_by('t1.userid')
$this->db->get()->result();
生成:
SELECT t1.*, t2.avg_rating, t3.*
FROM ratings t1
left join
(select product_id, avg(rating) as avg_rating from ratings group by product_id) t2
on t2.product_id=t1.product_id
left join users t3
on t1.user_id = t3.id
group by t1.user_id
并按预期输出.
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