Question

我的样本表如下：

Date            Product  Member_code
01/01/2015      ABC       1A
01/02/2016      BBB       1A
12/04/2013      SCC       2C
12/04/2014      AQQ       2C

预期结果

Member_code   Old_product       New_product
 1A             ABC               BBB
 2C             SCC               AQQ

感谢您的帮助欢呼声

Answer 1

捕获product列中最旧和最新值的最有效方法是使用first / last聚合函数。（个人观察：由于我不理解的原因，有许多从业人员，甚至是高级人员，他们都避免这种综合功能。）

使用Kaushik Nayak的设置（CREATE TABLE和INSERT声明）：

select   member_code,
         min(product) keep (dense_rank first order by p_date) as old_product,
         min(product) keep (dense_rank last  order by p_date) as new_product
from     yourtable
group by member_code
;

MEMBER_CODE  OLD_PRODUCT  NEW_PRODUCT
-----------  -----------  -----------
1A           ABC          BBB
2C           SCC          AQQ

Answer 2

一种方法是使用SELF JOIN。

SQL Fiddle

Oracle 11g R2架构设置：

CREATE TABLE yourtable
    (p_Date varchar2(10), Product varchar2(3), Member_code varchar2(2))
;

INSERT ALL 
    INTO yourtable (p_Date, Product, Member_code)
         VALUES ('2015-01-01', 'ABC', '1A')
    INTO yourtable (p_Date, Product, Member_code)
         VALUES ('2016-02-01', 'BBB', '1A')
    INTO yourtable (p_Date, Product, Member_code)
         VALUES ('2013-04-12', 'SCC', '2C')
    INTO yourtable (p_Date, Product, Member_code)
         VALUES ('2014-04-12', 'AQQ', '2C')
SELECT * FROM dual
;

查询1 ：

 SELECT t1.member_code,
       t1.product old_product,
       t2.product new_product
FROM   yourtable t1
       join yourtable t2
         ON t1.member_code = t2.member_code
            AND t1.p_date < t2.p_date

<强> Results ：

| MEMBER_CODE | OLD_PRODUCT | NEW_PRODUCT |
|-------------|-------------|-------------|
|          1A |         ABC |         BBB |
|          2C |         SCC |         AQQ |

根据日期将列转换为行oracle 1.5.5版

2 个答案: