MySQL乘子查询结果
问题描述：
我有一张看起来像的数据表
I have a table of data that looks like
+---------+-----------+------------+------------+
| 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
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);我想要做的简单版本是
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，因此取自
the more complex version I want to do is to weight the results based on the count from a u_id, so taking the results from
query 1
SELECT count FROM tablename WHERE u_id = 1 会返回
+-----------+------------+
| a_id | count |
+-----------+------------+
| 1 | 17 |
| 2 | 80 |
| 3 | 2 |
| 4 | 1 |
| 5 | 1 |
| 6 | 20 |
| 7 | 3 |
| 8 | 100 |
| 9 | 100 |
| 10 | 2 |
+-----------+------------+然后将用于计算总和的
+---------+-------------------+
| u_id | total |
+---------+-------------------+
| 1 | 83.15337423 |
| 3 | 65.05747126 |
| 2 | 1.704918033 |
+---------+-------------------+
例如，用u_id =3计算将通过
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
如何通过单个查询执行此操作?
How can I do this with a single query?
答
您可以使用子查询来执行此操作.获取特定ID的计数的查询是:
You can do this using a subquery. The query that gets the counts for a particular id is:
SELECT a_id, s_count FROM tablename WHERE u_id = <id>
您将需要将此子查询的结果保留到主表中，然后通过适当的乘法进行子处理，如下所示:
You will want to left join the result of this subquery into the main table, then sub over the appropriate multiplication, like so:
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