将两个复杂的SQL查询合并为一个

Question

我有3个表，其中一个是客户的表product_type_1和product_type_2。

客户

| id | name  |
|----|-------|
| 1  | Alex  |
| 2  | John  |
| 3  | Ahmad |
| 4  | Sam   |

product_type_1

| id | order_by  | Date
|----|--------------|-------|
| 1  |------ 1 ---- | 2019-03-01
| 2  |------ 2 ----| 2019-03-02
| 3  |------ 2 ----| 2019-03-03
| 4  |------ 3  ----| 2019-03-04

product_type_2

| id | order_by  | Date
|----|--------------|-------|
| 1  |------ 1 ---- | 2019-03-01
| 2  |------ 3 ----| 2019-03-02
| 3  |------ 3 ----| 2019-03-03
| 4  |------ 2  ----| 2019-03-04

最终输出将是特定年份每个月按客户名称分组的两种产品类型的总金额。我已经写过查询，但是一次只适用于一种产品类型。但是我想要两者的总和，即：

Customer | Jan | Feb | Mar .... Total<br>
:------------------------------------------------------:
John  ------   |  0  -- |--- 0   |--- 3  ......  3

John在2019年订购了3种产品。

查询为

select c.name,
               sum( month(o.order_date) = 1 and year(o.order_date)=2010) as Jan,
               sum( month(o.order_date) = 2 and year(o.order_date)=2010) as Feb,
               sum( month(o.order_date) = 3 and year(o.order_date)=2010) as Mar,
               sum( month(o.order_date) = 4 and year(o.order_date)=2010) as Apr,
               sum( month(o.order_date) = 5 and year(o.order_date)=2010) as May,
               sum( month(o.order_date) = 6 and year(o.order_date)=2010) as Jun,
               sum( month(o.order_date) = 7 and year(o.order_date)=2010) as Jul,
               sum( month(o.order_date) = 8 and year(o.order_date)=2010) as Aug,
               sum( month(o.order_date) = 9 and year(o.order_date)=2010) as Sep,
               sum( month(o.order_date) = 10 and year(o.order_date)=2010) as Oct,
               sum( month(o.order_date) = 11 and year(o.order_date)=2010) as Nov,
               sum( month(o.order_date) = 12 and year(o.order_date)=2010) as December,
               count(*) as total
        from customers c join
          (
          select order_by as cID, order_price , order_date
             from orders where year(order_date)=2010
          ) o
        on o.cID = c.id and o.order_price > 0
        group by c.name
        order by total desc

Answer 1

使用union all和聚合：

select c.id, c.name,
       sum( month(o.order_date) = 1 and year(o.order_date)=2010) as Jan,
       . . .
from customers c left join
     ((select order_by, date
       from product_type_1
      ) union all
      (select order_by, date
       from product_type_2
      ) 
     ) p12
     on p12.order_by = c.id
group by c.id, c.name