MySQLの乗算サブクエリの結果

Question

私は

 
CREATE TABLE IF NOT EXISTS tablename (u_id INT NOT NULL, a_id MEDIUMINT NOT NULL,s_count MEDIUMINT NOT NULL, weighted FLOAT NOT NULL)ENGINE=INNODB; 
INSERT INTO tablename (u_id,a_id,s_count,weighted) VALUES (1,1,17,0.0521472392638),(1,2,80,0.245398773006),(1,3,2,0.00613496932515),(1,4,1,0.00306748466258),(1,5,1,0.00306748466258),(1,6,20,0.0613496932515),(1,7,3,0.00920245398773),(1,8,100,0.306748466258),(1,9,100,0.306748466258),(1,10,2,0.00613496932515),(2,1,1,0.00327868852459),(2,2,1,0.00327868852459),(2,3,100,0.327868852459),(2,4,200,0.655737704918),(2,5,1,0.00327868852459),(2,6,1,0.00327868852459),(2,7,0,0.0),(2,8,0,0.0),(2,9,0,0.0),(2,10,1,0.00327868852459),(3,1,15,0.172413793103),(3,2,40,0.459770114943),(3,3,0,0.0),(3,4,0,0.0),(3,5,0,0.0),(3,6,10,0.114942528736),(3,7,1,0.0114942528736),(3,8,20,0.229885057471),(3,9,0,0.0),(3,10,1,0.0114942528736);

私が何をしたいの簡易版で再作成することができ

 
+---------+-----------+------------+------------+ 
| u_id | a_id  | count  | weighted | 
+---------+-----------+------------+------------+ 
|  1 |   1 |   17 | 0.0521472 | 
|  1 |   2 |   80 | 0.245399 | 
|  1 |   3 |   2 | 0.00613497 | 
|  1 |   4 |   1 | 0.00306748 | 
|  1 |   5 |   1 | 0.00306748 | 
|  1 |   6 |   20 | 0.0613497 | 
|  1 |   7 |   3 | 0.00920245 | 
|  1 |   8 |  100 | 0.306748 | 
|  1 |   9 |  100 | 0.306748 | 
|  1 |  10 |   2 | 0.00613497 | 
|  2 |   1 |   1 | 0.00327869 | 
|  2 |   2 |   1 | 0.00327869 | 
|  2 |   3 |  100 | 0.327869 | 
|  2 |   4 |  200 | 0.655738 | 
|  2 |   5 |   1 | 0.00327869 | 
|  2 |   6 |   1 | 0.00327869 | 
|  2 |   7 |   0 |   0 | 
|  2 |   8 |   0 |   0 | 
|  2 |   9 |   0 |   0 | 
|  2 |  10 |   1 | 0.00327869 | 
|  3 |   1 |   15 | 0.172414 | 
|  3 |   2 |   40 | 0.45977 | 
|  3 |   3 |   0 |   0 | 
|  3 |   4 |   0 |   0 | 
|  3 |   5 |   0 |   0 | 
|  3 |   6 |   10 | 0.114943 | 
|  3 |   7 |   1 | 0.0114943 | 
|  3 |   8 |   20 | 0.229885 | 
|  3 |   9 |   0 |   0 | 
|  3 |  10 |   1 | 0.0114943 | 
+---------+-----------+------------+------------+ 

よう

SELECT u_id, SUM(weighted) as total FROM tablename WHERE a_id IN (1,2,3,4,5,6,7,8,9) GROUP BY u_id ORDER BY total DESC;

で見えるデータのテーブルを持っています

結果が得られる

 
+---------+-------------------+ 
| u_id | total    | 
+---------+-------------------+ 
|  2 | 0.996721301227808 | 
|  1 | 0.993865059688687 | 
|  3 | 0.988505747169256 | 
+---------+-------------------+ 

私がやりたい、より複雑なバージョンがそう

 
query 1 
SELECT count FROM tablename WHERE u_id = 1

からの結果を取ることは次のようになり

 
+-----------+------------+ 
| a_id  | count  | 
+-----------+------------+ 
|   1 |   17 | 
|   2 |   80 | 
|   3 |   2 | 
|   4 |   1 | 
|   5 |   1 | 
|   6 |   20 | 
|   7 |   3 | 
|   8 |  100 | 
|   9 |  100 | 
|  10 |   2 | 
+-----------+------------+

を返す、U_IDからのカウントに基づいて結果を重み付けすることです合計を計算するために使用される、 woと計算する私は単一のクエリでこれを行うことができますどのように

sum(count value from query 1 * weighting value for u_id = 3 for each a_id) 

 
17 * 0.172413793 =2.931034483 
80 * 0.459770115 =36.7816092 
2 * 0   =0 
1 * 0   =0 
1 * 0   =0 
20 * 0.114942529 =2.298850575 
3 * 0.011494253 =0.034482759 
100 * 0.229885057 =22.98850575 
100 * 0   =0 
2 * 0.011494253 =0.022988506 
sums up to    65.05747126 

ことによって行うことがULD？

Answer 1

サブクエリを使用してこれを行うことができます。特定のIDのカウントを取得するクエリは、次のとおりです。

SELECT a_id, s_count FROM tablename WHERE u_id = <id> 

あなたは、適切な乗算の上にサブそして、メインテーブルには、このサブクエリの結果を結合する左のようにようにしたいと思うでしょう

：

SELECT u_id, SUM(counts.s_count * tablename.weighted) AS total FROM tablename 
LEFT JOIN (SELECT a_id, s_count FROM tablename WHERE u_id = 1) counts 
    ON tablename.a_id = counts.a_id 
GROUP BY u_id