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MySQL乘以子查询结果
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我有一个看起来像的数据表
+---------+-----------+------------+------------+ | 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 | +---------+-----------+------------+------------+
可以用它重新创建
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;
给出了结果
+---------+-------------------+ | 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,所以从中获取结果
query 1 SELECT count FROM tablename WHERE u_id = 1
会回来的
+-----------+------------+ | a_id | count | +-----------+------------+ | 1 | 17 | | 2 | 80 | | 3 | 2 | | 4 | 1 | | 5 | 1 | | 6 | 20 | | 7 | 3 | | 8 | 100 | | 9 | 100 | | 10 | 2 | +-----------+------------+
应该给出的,然后用它来计算总和
+---------+-------------------+
| u_id | total |
+---------+-------------------+
| 1 | 83.15337423 |
| 3 | 65.05747126 |
| 2 | 1.704918033 |
+---------+-------------------+
例如,使用u_id = 3进行计算
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
如何使用单个查询执行此操作？
解决方法:
您可以使用子查询执行此操作.获取特定id的计数的查询是：
SELECT a_id, s_count FROM tablename WHERE u_id = <id>
您将希望将此子查询的结果连接到主表中,然后转到相应的乘法上,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
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