堆栈内存溢出
  简体   繁体   English

Group By 组合到 CASE (SQL DB2)

[英]Group By combined to CASE (SQL DB2)

 原文  2021-03-29 08:54:12  9   1    sql/ db2

问题描述

I have a database of customers with this columns (sample):我有一个包含此列的客户数据库（示例）：

ID ID TIMESTAMP时间戳 TYP_EVEN TYP_EVEN NAME姓名 COUNTRY国家 GENDER性别 NBR_CHILDREN NBR_儿童 ADRESS地址 CUSTOMER_LINKED CUSTOMER_LINKED TYP_OF_LINK TYP_OF_LINK
044348547 044348547 2020-09-08-02.02.21.442908 2020-09-08-02.02.21.442908 0004 0004 NXXX CORINNE NXXX 科琳娜 FRA法兰克福机场 2 2 02 02 000000000 000000000
044379039 044379039 2020-07-17-11.17.55.410843 2020-07-17-11.17.55.410843 0013 0013 00 00 11 RUE XXXX XXX 11 街 XXXX XXX 000000000 000000000
044379039 044379039 2020-07-21-16.45.53.485200 2020-07-21-16.45.53.485200 0004 0004 KXXX STEPHANE KXXX 斯蒂芬妮 FRA法兰克福机场 1 1 00 00 000000000 000000000
044379039 044379039 2020-08-05-02.02.41.403053 2020-08-05-02.02.41.403053 0004 0004 KXXX STEPHANE KXXX 斯蒂芬妮 FRA法兰克福机场 1 1 00 00 000000000 000000000
044385911 044385911 2019-01-31-02.03.22.455660 2019-01-31-02.03.22.455660 0004 0004 SXXX PHILIPPE SXXX 菲利普 FRA法兰克福机场 1 1 02 02 000000000 000000000
044385911 044385911 2020-09-18-15.46.31.387588 2020-09-18-15.46.31.387588 0004 0004 SXXX PHILIPPE SXXX 菲利普 FRA法兰克福机场 1 1 02 02 000000000 000000000
044385911 044385911 2020-09-30-02.01.45.310449 2020-09-30-02.01.45.310449 0004 0004 SXXX PHILIPPE SXXX 菲利普 FRA法兰克福机场 1 1 02 02 000000000 000000000
044385911 044385911 2020-10-01-00.06.20.341427 2020-10-01-00.06.20.341427 0013 0013 00 00 19 RUE XXXX XXXX 19 街 XXXX XXXX 000000000 000000000
044680501 044680501 2021-03-10-02.02.42.979297 2021-03-10-02.02.42.979297 0004 0004 HXXX STEPHANE HXXX 斯蒂芬妮 FRA法兰克福机场 1 1 00 00 000000000 000000000
044680501 044680501 2021-03-10-02.02.42.980069 2021-03-10-02.02.42.980069 0013 0013 00 00 22 RUE XXXX XXXX 22 街 XXXX XXXX 000000000 000000000
044680501 044680501 2021-03-10-02.02.42.981083 2021-03-10-02.02.42.981083 0029 0029 00 00 044680519 044680519 MARIED已婚
044680519 044680519 2021-03-10-02.02.42.966522 2021-03-10-02.02.42.966522 0004 0004 HXXX LIDWINE HXXX 酒水 FRA法兰克福机场 2 2 02 02 000000000 000000000
044680519 044680519 2021-03-10-02.02.42.970031 2021-03-10-02.02.42.970031 0029 0029 00 00 044680501 044680501 MARIED已婚

TYP_EVEN : is type of event about the record for each customers ID, then OOO4 : is about informations of the customer (NAME, COUNTRY, GENDER, Number of children), 0013 : is about the ADRESS informations and 0029 : is about link beetwen customers TYP_EVEN : 是关于每个客户 ID 的记录的事件类型，然后OOO4 : 是关于客户信息 (NAME, COUNTRY, GENDER, Number of children), 0013 : 是关于 ADRESS 信息和0029 : 是关于链接 beetwen 客户

I would like to have one row by ID with all informations.我想按 ID 列出所有信息。 I proceed like this:我这样进行：

    SELECT
T1.ID,
CASE WHEN T1.TYP_EVEN = '0004' THEN T1.NAME END AS NAME,
CASE WHEN T1.TYP_EVEN = '0004' THEN T1.COUNTRY END AS COUNTRY,
CASE WHEN T1.TYP_EVEN = '0004' THEN T1.GENDER END AS GENDER,
CASE WHEN T1.TYP_EVEN = '0004' THEN T1.NBR_CHILDREN END AS NBR_CHILDREN,

CASE WHEN T1.TYP_EVEN = '0013' THEN T1.ADRESS END AS ADRESS,

CASE WHEN T1.TYP_EVEN = '0029' THEN T1.CUSTOMER_LINKED END AS CUSTOMER_LINKED,
CASE WHEN T1.TYP_EVEN = '0029' THEN T1.TYP_OF_LINK END AS TYP_OF_LINK
,T1.TIMESTAMP

FROM DB.CUSTOMERS T1

GROUP BY GROUP BY T1.ID,T1.TIMESTAMP, T1.TYP_EVEN,T1.NAME,T1.COUNTRY,T1.GENDER,T1.NBR_CHILDREN,T1.ADRESS
,T1.CUSTOMER_LINKED, T1.TYP_OF_LINK 
ORDER BY 1

the result should be:结果应该是：

ID ID NAME姓名 COUNTRY国家 GENDER性别 NBR_CHILDREN NBR_儿童 ADDRESS地址 CUSTOMER_LINKED CUSTOMER_LINKED TYP_OF_LINK TYP_OF_LINK
044348547 044348547 NXXX CORINNE NXXX 科琳娜 FRA法兰克福机场 2 2 02 02 11 RUE XXXX XXX 11 街 XXXX XXX 000000000 000000000
044379039 044379039 KXXX STEPHANE KXXX 斯蒂芬妮 FRA法兰克福机场 1 1 00 00 000000000 000000000
044385911 044385911 SXXX PHILIPPE SXXX 菲利普 FRA法兰克福机场 1 1 02 02 19 RUE XXXX XXXX 19 街 XXXX XXXX 000000000 000000000
044680501 044680501 HXXX STEPHANE HXXX 斯蒂芬妮 FRA法兰克福机场 1 1 00 00 22 RUE XXXX XXXX 22 街 XXXX XXXX 044680519 044680519 MARIED已婚
044680519 044680519 HXXX LIDWINE HXXX 酒水 FRA法兰克福机场 2 2 02 02 044680501 044680501 MARIED已婚

More information: if an ID have 2 rows with same TYP_EVEN then choose the latter one by the TIMESTAMP.更多信息：如果一个 ID 有 2 行具有相同的 TYP_EVEN，则通过 TIMESTAMP 选择后者。

Could someone have any idea to resolve this SQL Statement?有人可以解决这个 SQL 声明吗？

1 个解决方案

解决方案1
 1  2021-03-29 10:03:22

Here is a way to do what you want这是一种做你想做的事的方法

with customers (id, timestamp, typ_even, name, country, gender, nbr_children, adress, customer_linked, typ_of_link) as (
  values
  (044348547, timestamp('2020-09-08-02.02.21.442908'), 0004, 'NXXX CORINNE', 'FRA', 2, 02, NULL, 000000000, NULL),
  (044379039, '2020-07-17-11.17.55.410843', 0013, NULL, NULL, NULL, 00, '11 RUE XXXX XXX', 000000000, NULL),
  (044379039, '2020-07-21-16.45.53.485200', 0004, 'KXXX STEPHANE', 'FRA', 1, 00, NULL, 000000000, NULL),
  (044379039, '2020-08-05-02.02.41.403053', 0004, 'KXXX STEPHANE', 'FRA', 1, 00, NULL, 000000000, NULL),
  (044385911, '2019-01-31-02.03.22.455660', 0004, 'SXXX PHILIPPE', 'FRA', 1, 02, NULL, 000000000, NULL),
  (044385911, '2020-09-18-15.46.31.387588', 0004, 'SXXX PHILIPPE', 'FRA', 1, 02, NULL, 000000000, NULL),
  (044385911, '2020-09-30-02.01.45.310449', 0004, 'SXXX PHILIPPE', 'FRA', 1, 02, NULL, 000000000, NULL),
  (044385911, '2020-10-01-00.06.20.341427', 0013, NULL, NULL, NULL, 00, '19 RUE XXXX XXXX', 000000000, NULL),
  (044680501, '2021-03-10-02.02.42.979297', 0004, 'HXXX STEPHANE', 'FRA', 1, 00, NULL, 000000000, NULL),
  (044680501, '2021-03-10-02.02.42.980069', 0013, NULL, NULL, NULL, 00, '22 RUE XXXX XXXX', 000000000, NULL),
  (044680501, '2021-03-10-02.02.42.981083', 0029, NULL, NULL, NULL, 00, NULL, 044680519, 'MARIED'),
  (044680519, '2021-03-10-02.02.42.966522', 0004, 'HXXX LIDWINE', 'FRA', 2, 02, NULL, 000000000, NULL),
  (044680519, '2021-03-10-02.02.42.970031', 0029, NULL, NULL, NULL, 00, NULL, 044680501, 'MARIED')
)
select ID,NAME,COUNTRY,GENDER,NBR_CHILDREN,ADRESS,CUSTOMER_LINKED,TYP_OF_LINK from (
  select
    id,
    last_value(name, 'IGNORE NULLS') over(partition by id order by timestamp) name,
    last_value(country, 'IGNORE NULLS') over(partition by id order by timestamp) country,
    last_value(gender, 'IGNORE NULLS') over(partition by id order by timestamp) gender,
    last_value(nbr_children, 'IGNORE NULLS') over(partition by id order by timestamp) nbr_children,
    last_value(adress, 'IGNORE NULLS') over(partition by id order by timestamp ) adress,
    last_value(customer_linked, 'IGNORE NULLS') over(partition by id order by timestamp ) customer_linked,
    last_value(typ_of_link, 'IGNORE NULLS') over(partition by id order by timestamp ) typ_of_link,
    rownumber() over(partition by id order by timestamp desc) seq
  from customers
) where seq = 1
order by id

See db<>fiddle , Corinne NXXX has no address见db<>fiddle ，Corinne NXXX 没有地址

暂无
暂无

声明:本站的技术帖子网页，遵循CC BY-SA 4.0协议，如果您需要转载，请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

相关问题 与 group by 结合使用时的 SQL 案例 - SQL case when combined with group by DB2 SQL 不区分大小写 - DB2 SQL Case Insensitive DB2 SQL-GROUP BY的中位数 - DB2 SQL - median with GROUP BY DB2 Sql如何使用case语句？ - DB2 Sql how to use the case statement? 在Case DB2 SQL中使用JOIN - Using JOIN Within Case DB2 SQL 基于前一种情况的SQL DB2计数 - SQL DB2 count based on previous case 使用案例语句作为枢轴 - DB2 SQL - Using Case Statements As a Pivot - DB2 SQL 案例 sql 使用日期 DB2 - Case sql using dates DB2 SQL/DB2 替代扩展 CASE 语句 - SQL/DB2 Alternative to Extensive CASE Statements DB2 SQL更新中的案例结构 - CASE STRUCTURE IN A DB2 SQL UPDATE
相关标签
 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM