规范化 SQL 查询中的数据

Question

我有一个 SQL 查询 A（有关详细信息，请参阅下文），它返回如下表：

cluster  brand  amount
0         bos     600
0         phi     300
0         har     100
1         pro    2500
1         wal    1500
1         ash    1000
2         dil    4200
2         sor     500
2         van     300
...

但是，我想显示的不是数量，而是该数量与该集群中总量相比的比例，如下表所示：

cluster  brand  amount
0         bos    0.60
0         phi    0.30
0         har    0.10
1         pro    0.50
1         wal    0.30
1         ash    0.20
2         dil    0.84
2         sor    0.10
2         van    0.06
...

我应该如何更改我的 SQL 以便我可以访问一个集群中所有金额的总和，并且同一集群仍然有多个行？

** 详情**

SQL 服务器：MySQL，通过 python-MySQL 连接器连接。

用于生成第一个表的当前 SQL 查询：

SELECT c.cluster, brand, COUNT(o.id) AS brand_amount
FROM nyon_all.clustering AS c 
LEFT JOIN nyon_all.persons AS p ON c.pid = p.id 
LEFT JOIN nyon_all.orders AS o ON p.id = o.pid 
LEFT JOIN nyon_all.articles AS a ON o.aid = a.id 
LEFT JOIN nyon_all.brands AS ab ON a.brand_id = ab.id 
WHERE c.cluster_round = 'Org_2014-08-27_10:45:35'
GROUP BY cluster, brand 
HAVING brand_amount > 100
ORDER BY c.cluster ASC, brand_amount DESC;

表orders（主键id）链接persons（外键pid）和articles（外键aid）。 Articles有某个品牌（外键brand_id），与表brands中的一个名字相关。

可以使用以下 SQL 查询检索每个集群的文章总数：

SELECT c.cluster, COUNT(o.pid) AS amount
FROM nyon_all.clustering AS c 
LEFT JOIN nyon_all.persons AS p ON c.pid = p.id 
LEFT JOIN nyon_all.orders AS o ON p.id = o.pid 
WHERE c.cluster_round = 'Org_2014-08-27_10:45:35'
GROUP BY cluster
ORDER BY c.cluster ASC, amount DESC;

结果：

cluster amount
0        1000
1        5000
2        5000

但是，我似乎无法组合这两个 SQL 查询。

Answer 1

您可以对子查询进行连接，按集群对数量求和

select t1.cluster, amount / sumAmount 
from Table1 t1
join (select cluster, sum(amount) as sumAmount
      from Table1
      group by cluster)s
on t1.cluster = s.cluster

见SqlFiddle

编辑

SELECT 
    c.cluster, 
    brand, 
    COUNT(o.id) / coalesce(s.sumBrandAmount, 0) AS brand_amount -- of course it would be nice to check for dividing by 0
FROM nyon_all.clustering AS c 
LEFT JOIN nyon_all.persons AS p ON c.pid = p.id 
LEFT JOIN nyon_all.orders AS o ON p.id = o.pid 
LEFT JOIN nyon_all.articles AS a ON o.aid = a.id 
LEFT JOIN nyon_all.brands AS ab ON a.brand_id = ab.id 
LEFT JOIN (select c1.id, count(o1.id) as sumBrandAmount
           from nyon_all.clustering c1
           left join nyon_all.persons p1 on p1.id = c1.pid
           left join nony_all.orders as o1 on o1.id = p1.id
           --maybe some where clause as in your main query
           group by c1.id) s
                               ON s.id = c.id
WHERE c.cluster_round = 'Org_2014-08-27_10:45:35'
GROUP BY cluster, brand 
HAVING brand_amount > 100
ORDER BY c.cluster ASC, brand_amount DESC;