[英]Group By combined to CASE (SQL DB2)
我有一个包含此列的客户数据库(示例):
ID | 时间戳 | TYP_EVEN | 姓名 | 国家 | 性别 | NBR_儿童 | 地址 | CUSTOMER_LINKED | TYP_OF_LINK |
---|---|---|---|---|---|---|---|---|---|
044348547 | 2020-09-08-02.02.21.442908 | 0004 | NXXX 科琳娜 | 法兰克福机场 | 2 | 02 | 000000000 | ||
044379039 | 2020-07-17-11.17.55.410843 | 0013 | 00 | 11 街 XXXX XXX | 000000000 | ||||
044379039 | 2020-07-21-16.45.53.485200 | 0004 | KXXX 斯蒂芬妮 | 法兰克福机场 | 1 | 00 | 000000000 | ||
044379039 | 2020-08-05-02.02.41.403053 | 0004 | KXXX 斯蒂芬妮 | 法兰克福机场 | 1 | 00 | 000000000 | ||
044385911 | 2019-01-31-02.03.22.455660 | 0004 | SXXX 菲利普 | 法兰克福机场 | 1 | 02 | 000000000 | ||
044385911 | 2020-09-18-15.46.31.387588 | 0004 | SXXX 菲利普 | 法兰克福机场 | 1 | 02 | 000000000 | ||
044385911 | 2020-09-30-02.01.45.310449 | 0004 | SXXX 菲利普 | 法兰克福机场 | 1 | 02 | 000000000 | ||
044385911 | 2020-10-01-00.06.20.341427 | 0013 | 00 | 19 街 XXXX XXXX | 000000000 | ||||
044680501 | 2021-03-10-02.02.42.979297 | 0004 | HXXX 斯蒂芬妮 | 法兰克福机场 | 1 | 00 | 000000000 | ||
044680501 | 2021-03-10-02.02.42.980069 | 0013 | 00 | 22 街 XXXX XXXX | 000000000 | ||||
044680501 | 2021-03-10-02.02.42.981083 | 0029 | 00 | 044680519 | 已婚 | ||||
044680519 | 2021-03-10-02.02.42.966522 | 0004 | HXXX 酒水 | 法兰克福机场 | 2 | 02 | 000000000 | ||
044680519 | 2021-03-10-02.02.42.970031 | 0029 | 00 | 044680501 | 已婚 |
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
我想按 ID 列出所有信息。 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
结果应该是:
ID | 姓名 | 国家 | 性别 | NBR_儿童 | 地址 | CUSTOMER_LINKED | TYP_OF_LINK | ||
---|---|---|---|---|---|---|---|---|---|
044348547 | NXXX 科琳娜 | 法兰克福机场 | 2 | 02 | 11 街 XXXX XXX | 000000000 | |||
044379039 | KXXX 斯蒂芬妮 | 法兰克福机场 | 1 | 00 | 000000000 | ||||
044385911 | SXXX 菲利普 | 法兰克福机场 | 1 | 02 | 19 街 XXXX XXXX | 000000000 | |||
044680501 | HXXX 斯蒂芬妮 | 法兰克福机场 | 1 | 00 | 22 街 XXXX XXXX | 044680519 | 已婚 | ||
044680519 | HXXX 酒水 | 法兰克福机场 | 2 | 02 | 044680501 | 已婚 |
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。
有人可以解决这个 SQL 声明吗?
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
见db<>fiddle ,Corinne NXXX 没有地址
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