在购买另一个产品后找到购买另一个产品的客户

Question

我有一个业务逻辑问题，我不知道如何正确制定 sql 语句来检索它。我将简化业务模型以便于讨论。 我有一张表格，记录了哪些客户在哪一天购买了什么产品。

需要找出购买产品A后购买产品B的客户的实例。我提出了以下查询：

select t."Customer",t."Date" "Date2", t."Product" "Product2",a.*
from
(
select "Customer","Date", "Product"
from test2 where "Product"='A'
) a,
test2 t
where a."Customer"= t."Customer" and t."Product"='B' and t."Date">=a."Date"

这不太对，因为它过度计算了实例，因为如果我们查看表格，客户 1 在两次购买 A 后又购买了 B，但我的查询计数了 3 次

我圈出的实例不应该被计算在内。 有没有办法来解决这个问题？ 为方便起见，我附上了创建表 TEST2 的脚本。

  CREATE TABLE  "TEST2" 
   (    "Customer" VARCHAR2(26 BYTE), 
    "Date" DATE, 
    "Product" VARCHAR2(26 BYTE)
   );

Insert into  TEST2 ("Customer","Date","Product") values ('1',to_date('07-AUG-18','DD-MON-RR'),'A');
Insert into  TEST2 ("Customer","Date","Product") values ('1',to_date('07-AUG-18','DD-MON-RR'),'B');
Insert into  TEST2 ("Customer","Date","Product") values ('1',to_date('17-AUG-18','DD-MON-RR'),'A');
Insert into  TEST2 ("Customer","Date","Product") values ('1',to_date('27-SEP-18','DD-MON-RR'),'B');
Insert into  TEST2 ("Customer","Date","Product") values ('2',to_date('26-SEP-18','DD-MON-RR'),'A');
Insert into  TEST2 ("Customer","Date","Product") values ('3',to_date('01-OCT-18','DD-MON-RR'),'C');

Answer 1

假设您使用的是 12c 或更高版本，您可以使用 pattern matching (match_recognize) 执行此操作：

CREATE TABLE TEST2 (    
  customer      VARCHAR2(26 BYTE), 
  purchase_date DATE, 
  Product       VARCHAR2(26 BYTE)
);

Insert into TEST2 (customer,purchase_date,product) values ('1',to_date('07-AUG-18','DD-MON-RR'),'A');
Insert into TEST2 (customer,purchase_date,product) values ('1',to_date('07-AUG-18','DD-MON-RR'),'B');
Insert into TEST2 (customer,purchase_date,product) values ('1',to_date('17-AUG-18','DD-MON-RR'),'A');
Insert into TEST2 (customer,purchase_date,product) values ('1',to_date('27-SEP-18','DD-MON-RR'),'B');
Insert into TEST2 (customer,purchase_date,product) values ('2',to_date('26-SEP-18','DD-MON-RR'),'A');
Insert into TEST2 (customer,purchase_date,product) values ('3',to_date('01-OCT-18','DD-MON-RR'),'C');
commit;

select * from test2
match_recognize (
  partition by customer
  order by purchase_date, product
  measures
    proda.purchase_date prod_a_date,
    prodb.purchase_date prod_b_date
  pattern ( proda prodb )
  define
    proda as product = 'A',
    prodb as product = 'B'
);

CUSTOMER    PROD_A_DATE             PROD_B_DATE            
1           07-AUG-2018 00:00:00    07-AUG-2018 00:00:00    
1           17-AUG-2018 00:00:00    27-SEP-2018 00:00:00   

这说：

• 针对每位客户 (partition by customer)
• 按购买日期和产品排序 (order by purchase_date, product)
• 查找产品 A 的实例，然后是产品 B (pattern ( proda prodb ))
• define 部分说明产品是什么

模式是一个正则表达式。您可以重新定义变量以搜索您想要的任何产品（或使产品值绑定变量 - proda as product = :bindvar）。

如果您需要在 A 之后查找 B 的实例以及中间的其他产品购买，您可以使用排除语法 ( {-var-} )：

insert into TEST2 (customer,purchase_date,product) 
  values ('1',to_date('20-SEP-18','DD-MON-RR'),'C');

select * from test2
match_recognize (
  partition by customer
  order by purchase_date, product
  measures
    proda.purchase_date prod_a_date,
    prodb.purchase_date prod_b_date
  pattern ( proda {-otherprod*-} prodb )
  define
    proda as product = 'A',
    prodb as product = 'B',
    otherprod as product not in ( 'A', 'B' )
);

CUSTOMER    PROD_A_DATE             PROD_B_DATE            
1           07-AUG-2018 00:00:00    07-AUG-2018 00:00:00    
1           17-AUG-2018 00:00:00    27-SEP-2018 00:00:00  

Answer 2

您可以使用聚合和having 子句：

select customer
from test2
group by customer
having min(case when product = 'A' then date end) < max(case when product = 'B' then date end);

如果您想要日期，只需将having 中的函数添加到select。