Question

我试图找出一个查询，该查询将获得一个订单列表，其中包含产品A和B，C和D，X和Y等的组合。所以我需要一个列表来显示只有具有指定组合的订单。像下面的代码：

CREATE TABLE new
STORED AS PARQUET AS
SELECT DISTINCT
CASE t1.customer_number = RIGHT(t1.customer_number, LEN(t1.customer_number) - 2)
from Old;


customer_number     should become short_cust_no     
33764703        764703      
36764624        764624      
36763795        763795      
37764829        764829      
39766002        766002

Answer 1

如果您的表格数据

| ORDER_ID | PRODUCT_NAME | PRODUCT |
|----------|--------------|---------|
|        1 |        xxx01 |     A、B |
|        2 |        xxx02 |     B、C |
|        3 |        xxx03 |     C、D |
|        4 |        xxx04 |     D、X |
|        5 |        xxx05 |     E、Z |

试试这个脚本：

select order_id,product_name ,product
from orders 
where product like ('%A%B%') 
or product like ('%C%D%')

如果您的表格数据如此;

| ORDER_ID | PRODUCT_NAME | PRODUCT |
|----------|--------------|---------|
|        1 |        xxx01 |       A |
|        1 |        xxx02 |       B |
|        2 |        xxx03 |       C |
|        2 |        xxx04 |       D |
|        3 |        xxx05 |       E |

试试这个脚本：

select order_id
from orders 
group by order_id having   
  count( case when product in ('A','B') then 1 else null end  ) > 1
  or 
  count( case when product in ('C','D') then 1 else null end  ) > 1

Answer 2

让我将您的问题解释为需要特定的产品对。如果是，您可以使用group by和having。假设您的数据中没有重复项：

select o.order_id
from orders o 
group by o.order_id
having sum(case when o.product in ('A', 'B') then 1 else 0 end) = 2 or
       sum(case when o.product in ('C', 'D') then 1 else 0 end) = 2 or
       sum(case when o.product in ('X', 'Y') then 1 else 0 end) = 2;

如果您希望订单具有两个产品和那些货币对，请添加条件count(*) = 2.或者，如果您想要深奥，可以更改{{ 1}}类似于else。

选择具有产品A和B组合的订单

2 个答案: