MySQL 子查询结果相乘

我有一个数据表，看起来像

+---------+-----------+------------+------------+
| 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 |
+---------+-----------+------------+------------+

  

可以用以下命令重新创建

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);  

我想做的简单版本是

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;  

给出结果

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

我想要做的更复杂的版本是根据 u_id 的计数对结果进行加权，因此从

query 1
SELECT count FROM tablename WHERE u_id = 1   

会回来

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

然后用于计算总和，应该给出

+---------+-------------------+
| u_id    | total             |
+---------+-------------------+
|       1 | 83.15337423       |
|       3 | 65.05747126       |
|       2 | 1.704918033       |
+---------+-------------------+

例如计算u_id =3将由

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 
  

我如何通过单个查询来做到这一点？

您可以使用子查询来完成此操作。获取特定 id 的计数的查询是：

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

您需要将此子查询的结果左连接到主表中，然后对适当的乘法进行子运算，如下所示：

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