[英]SQL JOIN average value
我需要一些幫助來優化查詢
我有兩張桌子
表1 - users_location
; 表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 |
我需要按位置(例如AA)選擇用戶及其平均評分。 現在我的查詢是:
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
我認為SQL會計算所有用戶的平均評分,然后僅加入我需要的內容,這不是經過優化的。
對於mysql
您可以嘗試以下操作:
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';
這是sqlfiddle 。
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
根據Rick的反饋,以下是MySQL和SQL Server的答案:
MySQL的
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服務器
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]
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