Question

我有2个表，一个是供应商，一个是supplier_feedback - 如何计算每个供应商的平均评级？我目前有这个：

SELECT s.premium, s.supplier_id, s.name, s.phone, s.website, 
    s.price_low, s.price_high, s.address1, s.address2, s.town,
    s.county, s.postcode, 
    (SUM( f.rating ) / ( COUNT( f.rating ) -1 )) AS rate, 
    GROUP_CONCAT( REPLACE( t.name, ' ', ',' ) SEPARATOR ',' ) AS tags
FROM suppliers AS s
JOIN suppliers_to_tags AS st ON st.supplier_id = s.supplier_id
JOIN supplier_tags AS t ON t.tag_id = st.tag_id
JOIN supplier_feedback AS f ON s.supplier_id = f.supplier_id
GROUP BY s.supplier_id
HAVING tags LIKE '%HI%'
ORDER BY s.premium DESC
LIMIT 0 , 30

但是我得到了非常奇怪的结果，这肯定是不正确的。

supplier_feedback表将有许多记录，但第一条记录必须因各种原因打折。

Answer 1

标记上的连接会导致反馈行多次出现。这导致平均值偏离。您可以使用子查询为标记和平均值重写FROM部分。这确保了他们的计算不会干扰：

SELECT
     <other columns>
,    feedback.rating
,    suptags.tags
FROM suppliers AS s
JOIN (
    SELECT 
        st.supplier_id
    ,   GROUP_CONCAT(REPLACE( t.name, ' ', ',') SEPARATOR ',') AS tags
    FROM suppliers_to_tags AS st 
    JOIN supplier_tags AS t ON t.tag_id = st.tag_id
    GROUP BY st.supplier_id
) as suptags ON suptags.supplier_id = s.supplier_id
JOIN (
    SELECT 
        fb1.supplier_id
    ,   AVG(fb1.rating) as rating
    FROM supplier_feedback fb1
    WHERE fb1.feedback_id NOT IN (
        SELECT min(fb2.feedback_id) 
        FROM supplier_feedback fb2
        WHERE fb2.supplier_id = fb1.supplier_id
    )
    GROUP BY fb1.supplier_id
) feedback ON s.supplier_id = feedback.supplier_id

suptags子查询将每个供应商的所有标记添加到字符串中。 feedback子查询计算平均评分，不包括feedback_id最低的行。

Answer 2

使用子查询使事情更简单，这样就可以测试子查询是否与其他表的所有连接分开正确。

(SELECT supplier_id, AVG(rating) AS avg_rating 
FROM supplier_feedback WHERE feedback_id != x GROUP BY supplier_id)

假设feedback_id（无论反馈表的主键是什么）是不变的，您可以从子查询中排除要忽略的记录，其余的将被平均。

然后你可以加入：

    SELECT s.*, r.avg_rating
    FROM suppliers s
    JOIN 
    (SELECT supplier_id, AVG(rating) AS avg_rating 
    FROM supplier_feedback WHERE feedback_id != x GROUP BY supplier_id) r
    ON s.supplier_id = r.supplier_id

从另一个MySQL表计算平均值

2 个答案: