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MySQL平均连接表

转载 作者:行者123 更新时间:2023-12-01 00:49:12 25 4
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我有两个表:

沙龙排名:

+-----------+-----------+-----------+-----------+
| salon_id |category_id| user_id | rank |
+-----------+-----------+-----------+-----------+
| 10 | 1 | 999 | 1 |
| 10 | 2 | 999 | 1 |
| 10 | 1 | 888 | 5 |
| 10 | 2 | 888 | 5 |
+-----------+-----------+-----------+-----------+

类别:

+------+-----------+
| id | name |
+------+-----------+
| 1 | prices |
| 2 | service |
+------+-----------+

我正在尝试有效地获得如下所示的内容:

+-----------+-----------+-----------+
| salon_id |category_id| AVG(rank) |
+-----------+-----------+-----------+
| 10 | 1 | 3 |
| 10 | 2 | 3 |
+-----------+-----------+-----------+

对我来说最有意义的是这样的查询:

SELECT salon_id, category_id, AVG(rank)
FROM salon_ranks
INNER JOIN categories ON category_id = id
GROUP BY salon_id, category_id

但出于某种原因我没有得到平均值,而是得到这个:

+-----------+-----------+-----------+
| salon_id |category_id| AVG(rank) |
+-----------+-----------+-----------+
| 10 | 1 | 1 |
| 10 | 2 | 5 |
+-----------+-----------+-----------+

知道为什么会发生吗?什么可能是正确的查询?


对不起,这是我的错误。我从我的原始表中得到的数据是错误的。真的很抱歉误导你:(

最佳答案

您的解决方案:

http://www.sqlfiddle.com/#!2/ee5c6/2

SELECT SR.salon_id, SR.category_id, AVG(SR.rank)
FROM salon_ranks SR
INNER JOIN categories C ON SR.category_id = C.id
GROUP BY SR.salon_id, SR.category_id

关于MySQL平均连接表,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/17890215/

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