SQL JOIN平均值

Question

我需要一些帮助来优化我的查询

我有两张桌子

Table 1 - users_location; Table 2 - users_rating

+-------+---------+      +-------+--------+
|user   | location|      | user  | rating | 
+-------+---------+      +-------+--------+
| user1 |   AA    |      | user1 |   1    |
| user2 |   BB    |      | user1 |   3    |
| user3 |   CC    |      | user1 |   4    |
| user4 |   AA    |      | user2 |   3    |
| user5 |   BB    |      | user4 |   2    |
| user6 |   CC    |      | user4 |   4    |

我需要 select 位用户及其按位置划分的平均评分（例如 AA）。现在我的查询是：

SELECT * 
FROM users_loaction AS TABLE1
LEFT JOIN (
    SELECT  
        user, 
        AVG(rating) 
    FROM users_rating group by user
) AS TABLE2 
ON  TABLE1.user = TABLE2.user
WHERE TABLE1.location = AA

我认为SQL会计算所有用户的平均评分，然后只加入我需要的，这不是优化。

Answer 1

http://sqlfiddle.com/#!9/c34c8d/1

SELECT t.*, AVG(t2.rating) 
FROM users_location t
LEFT JOIN  users_rating t2
ON  t.user = t2.user 
WHERE t.location = 'AA'
GROUP BY t.Location;

SELECT t.*, AVG(t2.rating) 
FROM users_location t
LEFT JOIN  users_rating t2
ON  t.user = t2.user 
WHERE t.location = 'AA'
GROUP BY t.user;

SELECT t.*, AVG(t2.rating) 
FROM users_location t
LEFT JOIN  users_rating t2
ON  t.user = t2.user 
WHERE t.location = 'AA'
GROUP BY t.Location, t.user

Answer 2

根据 Rick 的反馈，以下是 MySQL 和 SQL 服务器的答案：

MySQL

http://sqlfiddle.com/#!9/9e9be8/1

SELECT l.user, AVG(r.rating) FROM users_location l
    LEFT JOIN users_rating r ON l.user = r.user
    WHERE l.location = 'AA'
    GROUP BY l.user

SQL 服务器

http://sqlfiddle.com/#!6/379e6/1

SELECT l.[user], AVG(r.rating) FROM users_location l
    LEFT JOIN users_rating r ON l.[user] = r.[user]
    WHERE l.location = 'AA'
    GROUP BY l.[user]

Answer 3

对于 mysql 你可以试试这个：

SELECT *, 
    (SELECT AVG(rating) FROM users_rating AS ur WHERE ur.user = ul.user ) AS rating
FROM users_location AS ul
WHERE ul.location = 'AA';

这里是sqlfiddle。

SQL JOIN平均值

SQL JOIN average value

mysql

sql-server

optimization

performance

query-optimization