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。为了简化讨论，我将简化业务模型。我有一张桌子，记录着哪些顾客在哪一天购买了什么产品。

需要找出购买了产品A后购买了产品B的客户实例。我想到了以下查询：

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"

这不是很正确，因为它已经对实例进行了过多计数，因为如果我们看表，客户1在2次购买A之后就购买了B，但我的查询已经计算了3次

我圈出的实例不算在内。有没有办法来解决这个问题？为了方便起见，我附加了一个脚本来创建表TEST2。

  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

假设您的年纪在12c或更高，则可以使用pattern matching (match_recognize)：

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：

对于每个客户（partition by customer）
按购买日期和产品（order by purchase_date, product）排序
查找产品A的实例，然后立即查找产品B（pattern ( proda prodb )）
define部分说明了什么产品

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

如果您需要在购买其他产品之后找到A之后的B实例，则可以使用排除语法（{-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);

如果需要日期，只需将having中的函数添加到select。

在购买其他产品后找到购买产品的客户

2 个答案: