嵌套查询 - 寻找更好的解决方案

Question

考虑一个销售应用程序,我们有两个表SALES和INTERNAL_SALES.

• 表"销售额"表示公司外部每个销售人员进行的交易数量.
• 表"INTERNAL_SALES"表示公司内每个销售人员与另一个销售人员进行的交易数量.

销售表:即使交易为零,每个日期对每个销售人员也有一个条目.

id |  day          | sales_person | number_of_transactions
1  | 2011-08-01    | Tom          | 1000
2  | 2011-08-01    | Ben          | 500
3  | 2011-08-01    | Anne         | 1500
4  | 2011-08-02    | Tom          | 0
5  | 2011-08-02    | Ben          | 800
6  | 2011-08-02    | Anne         | 900
7  | 2011-08-03    | Tom          | 3000
8  | 2011-08-03    | Ben          | 0
9  | 2011-08-03    | Anne         | 40

INTERNAL_SALES表:此表仅记录销售人员之间实际进行的交易.

id | day        | sales_person_from | sales_person_to | number_of_transactions
 0 | 2011-08-01 | Tom               | Ben             | 10
 1 | 2011-08-01 | Tom               | Anne            | 20
 2 | 2011-08-01 | Ben               | Tom             | 50
 3 | 2011-08-03 | Anne              | Tom             | 30
 4 | 2011-08-03 | Anne              | Tom             | 30

现在问题是每个销售人员每天提出总交易.我这样做的方式是:

SELECT day, sales_person, sum(num_transactions) from
(
  SELECT day, sales_person, number_of_transactions As num_transactions FROM sales;
  UNION
  SELECT day, sales_person_from As sales_person, sum(number_of_transactions) As num_transactions FROM internal_sales GROUP BY day, sales_person_from;
)
GROUP BY day, sales_person;

这太慢了,看起来很难看.我正在寻求更好的解决方案.顺便说一下,在Oracle中使用的数据库和我无法控制数据库,除了我可以对它运行查询.

Answer 1

不需要聚合两次,并且union运算符通常执行隐式唯一排序,这在您的情况下也是不必要的.

SELECT day, sales_person, sum(num_transactions) from
(
    SELECT day, sales_person, number_of_transactions As num_transactions FROM sales;
    UNION ALL 
    SELECT day, sales_person_from, number_of_transactions FROM internal_sales;
)
GROUP BY day, sales_person;

删除中间聚合和唯一排序应该有所帮助.